i

Size: px
Start display at page:

Download "i"

Transcription

1 mailto:

2 i

3 { : : : : : : : : : : : : : : : : : : : : : : : : 1 12 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8 22 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8 23 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9 24 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 45 ii

4 iii 56 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :

5 1 11 { N H 2 O (1) (2) 1

6 ! ' ! ' 2: ! 52! (26!) 2 (26!) 2 =52!=1= ! = {

7 (1) 1 u = f(x) x f(x + h) 0 f(x) lim h!0 h x df dx ; du dx ; f 0 (x) 2 u = f(x; y) y f 1 x f @x =lim h!0 f(x + h; y) 0 f(x; ; f x(x; y) (1:1) y 1 f(x; y) = (x; y) @f f @x@y f(x; 2 f (2) H

8 4 1 x u = f(x) du dx = f 0 (x) du = f 0 (x)dx (1:4) u x 4x; 4u 4u ' f 0 (x)4x du dx f(x + dx) =f(x)+f 0 (x)dx (1:5) d(cu) =cdu ; d(uv) =vdu + udv ; dy = dy dx (1:6) dx x u x u x u x dy =0 du u = f(x; y) dy = f x(x; y) dx + f y (x; y) dy (1:7) (3) 1 du = A(x) dx (1:8) A(x) Z x f(x) = A(x 0 )dx 0 (1:9)

9 12 5 u x u = f(x) (1:10) 2 du = A(x; y; :::) dx + B(x; y; :::) dy + ::: (1:11) f(x; y; :::) x y; ::: u u = f(x; y; :::) (111) du xdx + ydy = d(x 2 =2+y 2 =2) ; ydx + xdy = d(xy) 0ydx + xdy @x (1:12) ( 2 f=@x@y 100% (x; y; :::) du A(x; y; :::) B(x; y; :::) = dx + dy + ::: (1:13) (x; y; :::) (x; y; :::) S(x; y; :::) (x; y; :::) arctan tan 0ydx + xdy x 2 + y 2 = d(arctan y x ) (1:14)

10 6 1 (4) u = f(x; y) x; y u u = u(x; y) u = f(x; y) y x z y(x; z) u y x; z u = f(x; y(x; z)) = g(x; z) u x f x u z y x z x x y x (5) 3 x; y; z '(x; y; z) =0 (1:17) dz =0 (1:18) x (y;z) (@x=@y) z z dz =0 dx dy @x z y z z x x; @y z z y etc: = 01

11 12 @y @z z y = = y y x 1 (i) f x (x; y + h) 0 f x (x; y) = lim h!0 h 1 = lim lim h!0 h 0!0 h 1 = lim lim h 0!0 h!0 f(x + h 0 ;y+ h) 0 f(x; y + h) h 0 0 f(x + h0 ;y) 0 f(x; y) h 0 f(x + h 0 ;y+ h) 0 f(x + h 0 ;y) f(x; y + h) 0 f(x; y) h 0 0 h h = lim h 0!0 f y (x + h 0 ;y) 0 f y (x; y) h (120) du = f(x + dx; y + dy) 0 f(x; y) = f(x + dx; y + dy) 0 f(x; y + dy)+f(x; y + dy) 0 f(x; y) = f x (x; y + dy) dx + f y (x; y)dy = h f x (x; dyi dx + f y (x; y)dy dxdy (@ 2 f=@x@y)dxdy dx + dy = dx @x dz @' dz

12

13 23 9 A = C ; B = C A = B (2:1) Pa ( ) V P V = V (;P) F (;V;P)=0 (2:2) (;V;P) dv = d + P = P (2:4) = 0 (2:5)

14 10 2 d 0 1 dv 0 dp =0 (2:6) = V = = P (2:8) 24 Boyle PV = (2:9) PV = R (2:10) R, 1 (V;) V = R=P V! 0-1 R (= PV=) P

15 :16 (2:11) : K 600K 0 N A =6: ) R =8:31 [J=mol 1 K] (2:12) MKS N/m 2 Pa m 3 J R R =1 MKS 3 300K 600K =1=; =1=P; = P= PV 10 Pa 2 K 3 A MKS m, kg, s

16 3 18 caloric 1 31 A B U A B W (A! B) = U(B) 0 U(A) (3:1) A B V U(;V ) d 0 W = du (3:2) 1 12

17 32 13 U d 0 32 W (A! B) 6= U(B) 0 U(A) (3:3) [U(B) 0 U(A)] 0 W (A! B) = Q(A! B) (3:4) W (A! B) + Q(A! B) = U(B) 0 U(A) (3:5) d 0 W + d 0 Q = du (3:6) d 0 W d 0 Q

18 14 3 d 0 Q =0 d 0 W = du d 0 W =0 d 0 Q = du 95 (Joule) cal 15 = 4:1855 J M 1cal 1g cal 15 cal J cal J

19 p35 (i) (ii) F P A F = PA dx dv = 0Adx d 0 W = Fdx= 0PdV (3:7) A P F dx P V W V 1 V 2 d 0 W<0 (P; V ) P V P -V V - 0 0

20 16 3 V V 0 PdV = R dv=v = R d(log V ) W = R log(v 0 =V ) d 0 W = 0PdV d 0 W = d(0r log V ) 34 1K d 0 Q = du + PdV (3:8) U (;V) U = du = d + V dv =0 C V =[d 0 Q=d ] dv =0 =[du=d ] dv V (3:10) / d 0 Q = du + PdV = C V d + P i dv (3:11) dp =0 C P = [d 0 Q=d] dp =0 = C V + + h = C V + P + P i (312) H = U + PV du = dh 0 PdV 0 VdP (3:13) d 0 Q = du + PdV = dh 0 VdP (3:14)

21 35 17 C P =[d 0 Q=d ] dp =0 =[dh=d ] dp P (3:15) d 0 Q d 0 Q = du d 0 Q d 0 Q = dh C V ;C P d 0 Q = C V d + C P 0 C V V dv (3:16) d 0 Q =0 (;V ) C V d + C P 0 C V V dv =0 (3:17) ad adiabatic ad = 0V 01 [dv=dp ] d 0 Q=0 ad =(C V =C P ) [ d dv dv = V d 0 V dp dv dp ] 6 C P >C V > 0 > ad > 0 (V;P) P -V V ) 35 PV = =0 GayLussac

22 V d 0 W =0 d 0 Q =0 V dv =0 (3:19) =0 (3:20) 2 C V =(@U=@) V H = U + PV = U()+R (115) C P C P ( ) 0 C V ( )=R (3:21) C V =3R=2 d 0 Q = du + PdV = C V ()d + PdV =0 (3:22) P = R=V dv V = 0C V ( ) R d (3:23) log V 2 0 log V 1 = 0 Z 2 1 C V () R d (3:24) 2

23 36 19 " 8( )=exp 0 Z CV ( 0 ) R 0 d 0 # (3:25) V 0 V = 8( 0 ) 8() (3:26) = C P =C V (3:27) R=C V =(C P 0 C V )=C V = 0 1 dv V = 0 1 d 0 1 (3:28) ( 0 1) log V + log = constant V 01 = (3:29) PV = R PV = (3:30) 36 W (A! A) + Q(A! A) = U(A) 0 U(A) =0 (3:31) Q =0 (0W) > 0

24 20 3 Carnot 1 1 A B 2 B C 3 C D 0 4 D A p16 p19 1 jw 1 j = Q = R log(v B =V A ) 2 jw 2 j = U( ) 0 U( 0 ) V C =V B =8( 0 )=8( ) 3 W 3 = jq 0 j = R log(v C =V D ) 4 W 4 = U( ) 0 U( 0 )(=jw 2 j V D =V A =8( 0 )=8( ) P A : 1 B 4 2 D 3 0 : C 0 V 3 = I PdV = jw 1 j + jw 2 j0w 3 0 W 4 = Q 0jQ 0 j (332) V C =V B =8( 0 )=8( )=V D =V A V B =V A = V C =V D I 3 Q jq 0 j = R log(v B=V A ) R 0 log(v C =V D ) = 0 (3:33) Q = jq0 j 0 (3:34)

25 36 21 = (3:35) 0 = Q 0jQ0 j Q =10 0 (< 1) (3:36) 100% 90%

26 4 100% 41 (1) (2) (3) (4) 42 22

27 Clausius 44 homson Kelvin 1 1

28 24 4 Ostwald 100% 1 C 1 Q W (= Q) A 2 C A 1 2 Q 2 A C W (= Q>0) Q>0 C 2 Q + A Q> C' 1 2 Q Q B W

29 45 25 Q 0 W 1 C' B 2 1 Q 0 (Q 0 W )=W Q 2 C' Q>0 Q B! W (> 0) Q0 (= Q 0 W ) C' 2 Q Q B W Q 0 Q 0 (= W ) > C C C C 1) 2) 3) (a) (b)

30 26 4 C (;V ) (;2V ) V 2V if exists x p

31 % 48 2 Q 2 1 Q 1 ( ) Q 2 = f( 2) Q 1 f( 1 ) (4:1) p34 f( ) / 36 Q 2 = Q 1 (4:2) ; 1 ;Q 2 ;Q 1 ; 0 ;Q;jQ 0 j =273:16 2 (336) 0 = Q 2 0 Q 1 Q 2 = (4:3) Q 1 (42)

32 28 4 Q 1 0Q 1 ( ) Q Q 2 2 = 0 (4:4) f i g fq i g p35 X i d 0 Q I d 0 Q =0 (4:6) ( ) ( ) Q i i =0 (4:5) 0 P V O A Z A d 0 Q O( ) = S A (4:7) A S A O I d 0 Q = Z d 0 Q Z + d 0 Q =0 (4:8) Z d 0 Q Z = 0 d 0 Q (4:9) Z d 0 Q Z = d 0 Q (4:10)

33 48 29 S A B Z B A( ) d 0 Q = S B 0 S A (4:11) d 0 Q = ds (4:12) d 0 Q 1 p5 d 0 Q = ds (4:13) d 0 W = 0PdV d 0 Q =0 A Q = I ds B ds =0 (4:14) D C S 0 (S; ) S- = I C ds (4:15) 48 p35 (45) W = 0 P i Q 0 i > 0 X i Q i i 0 (4:16) 2 Q 2 1 jq 1 j Q jq 1j 1 0 (4:17)

34 > 1 = Q 2 0jQ 1 j Q = 0 (4:18) W 1 Q 1 (> 0) Q 2 = Q 1 + W 2 Q 1 + (0Q 2) = Q 2 0 W 0 Q Q 2 W 1 < Q 2 W K 10 20K I d 0 Q ext 0 (4:19) ext d 0 Q = ext ext d 0 Q ds = d 0 Q p36 C A B A I d 0 Q ext = Z B A d 0 Q ext + Z A B( ) d 0 Q = Z B A d 0 Q ext +[S A 0 S B ] 0 (4:20) Z B A d 0 Q ext S B 0 S A (4:21) d0 Q ext ds (4:22)

35 (422) d 0 Q =0 ds 0 (422) d 0 Q ext = 0dS ext = (4:23) d(s + S ext ) 0 (4:24) d =0 ds = d 0 Q = du( )+PdV = C V ( )d + R dv V (4:25) ds = R dv V = Rd log V 4S = R Z V 0 V dv V = R log V 0 V (4:26) V 0 >V 4S >0 4S >0 4S <0

36 32 4 C 1 ; 2 f C( f 0 1 )=C( 2 0 f ) f = (4:27) f d 0 Q i = Cd ds i = d 0 Q= = Cd= 4S i = Z f i Cd 4S = 4S 1 + 4S 2 = C log f 1 + C log f 2 = C log f i (4:28) = C log 2 f 1 2 > 0 (4:29) ( )=2 > p (1) A(x; y);b(x; y) d 0 Q = A(x; y) dx + B(x; y) dy (4:30) d 0 Q A(x; y) = (x; y) (x; y) dx + B(x; y) (x; y) dy (4:31) (x; y) 12,p5 d 0 Q =0 (x; y) dy A(x; y) = 0 dx B(x; y) (4:32) S(x; y) = constant (4:33) dy =0 (4:34)

37 d 0 Q =0 A(x; = (4:35) (x; y) A B d 0 dy = ds (4:36) (2) d 0 Q = X i A i (x 1 ;x 2 ;:::) dx i =0 (4:37) S(x 1 ;x 2 ; :::) = constant (4:38) d 0 Q = X i A i (x 1 ;x 2 ; :::) dx i =0 (4:39) ds =0 Caratheodory Q>0 jw j > 0

38 =( )= 2 1 =0 100% ) Q 1 ;Q 2 1 ; 2 C W W C,C' C W C' C 0 Q 0 2 = Q 2 Q 0 1 = Q 1 Q 2 0 Q 0 2 < 0 Q 2 0 Q 0 2 > 0 ] C Q 2 Q 1 2 W 1 C 0 Q 0 2 Q 0 1 (2) Q 2 =Q 1 W n Q 2 ( 1 ; 2 ;nw)=nq 2 ( 1 ; 2 ;W) Q 1 ( 1 ; 2 ;nw)=nq 1 ( 1 ; 2 ;W) n n Q 2 =Q 1 W 2 ; 1 Q 2 Q 1 = F ( 2 ; 1 ) (3)F ( 2 ; 1 )=f( 2 )=f( 1 ) 2 1 C,C' C C' Q C C' C Q 2 =Q = F ( 2 ;) C' Q=Q 1 = F (; 1 ) C+C'+ Q 2 =Q 1 = F ( 2 ; 1 ) F ( 2 ;) F (; 1 )=F ( 2 ; 1 ) 2 C Q Q C' Q 2 Q 1 F ( 2 ; 1 )=f( 2 )=f( 1 ) 1

39 (45) C i Q i i C i Q 0 i i Q i C i Q i i = Q0 i X i Q i i = 1 X i Q 0 i (3) i Q i C f i g fc i g P i Q0 i W 0 = W + P i W i = P i Q i + P i (Q0 i 0 Q i)= P i Q0 i W 0 > 0 W 0 < 0 W 0 =0 (3) P i Q i= i =0 1 2 n C C 1 C 2 C n Q 1 Q 2 Q n W Q 0 1 Q 1 W 1 Q 0 2 Q 2 W 2 Q 0 n Q n W n ( )

40 36 4 (1) (2) (3) (4) (P;V ) 10 (P; V ) PV = R 4S = R log 2 > 0 (P; V ) (P=2; 2V ) (4:40) (422) d 0 Q ds (4:41) (413)

41 5 51 (0) S(U; V ) d 0 Q = du 0 d 0 W ;P d 0 W = 0PdV; dq = ds ds = du +PdV (5:1) (U; V ) ds = 1 du + (1) U(S; V ) du = ds 0 PdV = (2) H(S; P) V V = U = P = 0P S H = U + PV (S; P ) dh = du + PdV dh = ds + VdP = ; = (LeGendre) (x; y) L(x; y) P dl = X(x; y)dx + Y (x; y)dy (5:5) x X(x; y) = ; dx = L xx dx + L xy dy y 37 S

42 38 5 x (X; y) dx = dx + dy (= 1 dx 0 L xy L xx L xx y X dl = Xdx + h = X dx + y = X L xx @x dx y h 0 L xy L xx X + Y i X + Y i dy X i dy + Ydy dy (58) L 0 = L 0 xx (5:9) dl 0 = dl 0 xdx 0 Xdx dl 0 = 0xdX + Y (x(x; y = 0x X = Y (5:10) (3) (Helmholtz) (S; V )! (;V ) ; F = U 0 S df = 0Sd V = = 0P (5:11) (4) (Gibbs) (S; V )! (;P) ; G = U 0 S + PV dg = 0Sd P = = V (5:12) { { d =0 df = 0PdV = d 0 W U = F + S F S A B F A B W (A! B) = F (B) 0 F (A) (5:13) d 0 W = df (5:14)

43 52 39 F U U = F + S = F 0 = 0 2 @(G= ) H = G + S = G 0 = 0 p83 P P dl = A(x; y; :::)dx + B(x; y; :::)dy L etc: (5:18) (1) du = ds 0 PdV! = 0 @V (2) dh = ds + VdP! = @P (3) df = 0Sd 0 PdV! @V (4) dg = 0Sd + VdP! = (Maxwell) V = V (;P) P

44 40 5 du = ds = = 0 P V V (523) V = f(p=) V = P= C P >C V C P 0 C @P V @ P P 2 P P = V2 (> 0) (524) >0 62 C V =[d 0 Q=d] dv =0 =[ds=d ] dv =0 = C P =[d 0 Q=d] dp =0 =[ds=d ] dp =0 = @S @ 2 P P = f 1 (V ) + f 2 (V ) C V

45 U =0 (5:27) 12 (5) @U V = 01 (5:28) (i) (@V=@U) =1=(@U=@V ) (@U=@) V = U CV = h V i CV = P 0 C V (5:29) =(@P=@) V - - P 1 1 P 2 2 ( ) P 1 ; 1 P 2 ; 2 1 P 1 V 1 P 2 V 2 P 1 V 1 0 P 2 V 2 (=U 2 0 U 1 (5:30) U 2 0 U 1 U 1 + P 1 V 1 = U 2 + P 2 V 2 H 1 = H 2 H (5:32)

46 42 @H = @ H = H i C P = 0 V C P P 0 = V ( 0 1) C P (5:35) = V 01 (@V=@) P =0 >0 <0 200K >0 4P <0 4 <0 - dh = ds+vdp =0 ds = 0(V=)dP = 0R(dP=P ) 4S = 0R log(p 2 =P 1 ) - p

47 54 43 { (Planck) (Einstein) 19 (a) U=V = u() (b) P = u=3 @P = 0 P = 0 P @ 3 d 0 u (5:36) = U V = u (5:37) du u =4d (5:38) log u = constant + 4 log u = 4 ; C V =4 3 ( ) (5:39) u c c=4 93 K = c 4 u = 4 (5:40)

48 (Stefan-Boltzmann) - =5: [W=m 2 K 4 ] (5:41) W = 25 k B 4 15c 2 h 3 (5:42) k B = R=N A h ds = du + PdV = d( 4 V )+ 4 3 dv =4 3 Vd dv (5:43) ds =4 2 Vd dv = d 4 3 V 3 (5:44) S V S = 4 3 V 3 ; F = U 0 S = 0 4 V 3 (5:45) 1 10mm 1m 25kW 1450K K - D R 200 1m km 30% H 2 O 10

49 (1) A l V = Al n n J H = nj B 8=AB = 0nl d8 dt = 0nV db dt (5:46) dq(= Jdt) (0)Jdt = n(jdt)v db dt = VHdB (5:47) V =1 0 H 2 d = 0 HdH (5:48) 2 d 0 W = HdB 0 0 HdH = Hd(B 0 0 H)=HdM (5:49) M = B 0 0 H du = ds + HdM df = 0Sd + HdM dg = 0Sd 0 MdH F @H = @ @M + H + H = 0 M (5:51)

50 46 5 M = f(h=) (Curie) M = 0 ( )H ; 0 = C (5:52) @M ds = d + dh = C H d + H H (5:53) CH (5:54) (@M=@) H < 0 (2) x f dx fdx S du = ds + fdx ; df = 0Sd + fdx = x H (5:56) f (a) x (@f=@x) > 0 (b) f= = = + f + f = 0 2 =0 x = ds = 0d @f = @ f x

51 < 0 = 0 = = > @ C x S x (3) ( 7 F (; ) 1 2 ( ) da d 0 W = df = 0Sd + ( )da ; = ( ) F=A =(@F=@A) = () ( )=F=A = (5:62) x x df = 0Sd + da U = F + S = 0 = 0A A d ; S A = 0 d d h U = F + S = A 0 d i U d ; = ( ) 0 d A d (5:63) (5:64)

52 (1) PV = f( ) (2) (@U=@V ) =0 f( ) / (3) (Dalton) fn i g PV i = n i R (5:65) V = P i V i PV = P X i V i = X i n i R = nr n = X i n i (5:66) V P i = n i R=V V = X i V i = X i n i R P = X i P i P V ; P = X i P i (5:67) P P i (4) C V ( )= du Z d U( )=U 0 + C V ( 0 )d 0 (5:68) 0 C P ( )= dh d H( )=H 0 + C P ( 0 )d 0 (5:69) Z C P ( 0 ) C P ( )= S(;P)=S 0 (P P 0 0 d = 0 = P ; S 0(P) =S 0 0 R log P (5:71) P 0 G(;P) = H 0 S P S(;P)=S 0 0 R log P P 0 + S(;V )=S 0 + R log V V 0 + = H 0 0 S 0 + Z 0 Z Z 0 C P ( 0 ) 0 d 0 (5:72) Z C P ( 0 )d 0 + R log P P C V ( 0 ) 0 d 0 (5:73) Z 0 C P ( 0 ) 0 d 0 (574)

53 n n u i (x 1 ;x 2 ; ::::; x n ) (i =1; 2;:::;n) (5:75) (Jacobian) 1 ;u 2 ; :::; u n 1 ;x 2 ; :::; x n ) = i o j f@u i =@u j i =@u j ij- [ fu i g fy i g fx i 1 ;u 2 ; :::; u n 1 ;x 2 ; :::; x n ) 1;u 2 ; :::; u n 1 ;y 2 ; :::; y n 1 ;y 2 ;:::;y n 1 ;x 2 ;:::;x n ) i k k=1 j A; B det(ab) = (deta)(detb) u = u(x; y; y;z;::: y; z; :::) u 1 = u; u 2 = y;u 3 = z; :::; x 1 = x; x 2 = y;x 3 @u=@z :::::::: :::::::: :::::::: ::::: ::::: ::::: ::::: ::::: ::::: ::::: ::::: ::::: @(V; ) = S = = C P C P

54 d 0 Q ext ds (6:1) ext d 0 Q (1) d 0 Q =0 ds 0 (6:2) 2 S = (6:3) d 0 Q = du 0 d 0 W (61) du 0 d 0 W ext ds (6:4) (2) ds =0 du d 0 W (2-1) d 0 W = 0PdV =0 du 0 (6:5) U = (6:6) 50

55 62 51 (3) ext = (64) du 0 d 0 W ds d(u 0 S)=dF d 0 W (6:7) d 0 W<0 jd 0 WjjdF j (3-1) df =0 d 0 W 0 (6:8) d 0 W = 0PdV (67) df + PdV = dg 0 VdP 0 (6:9) (3-2) d =0;dV =0 df 0 (6:10) F = (6:11) (3-3) d =0;dP =0 dg 0 (6:12) G = (6:13)

56 52 6 f(x 1 ;x 2 ; :::) x 0 =(x 10 ;x 20 ; :::) x x 0 x =(x 1 ;x 2 ;:::) 0 x i + 1 f 0 x i x j + i j f = X i i;j = (1) f + (2) f + ::::::: (614) 0 x 0 (1) f =0 ; (2) f>0 (6:15) (2) f<0 x dx { { dx S = S 1 (U 1 ;V 1 )+S 2 (U 2 ;V 2 ) V 1 ;V 2 S = U 1 + U 2 V1 2 V2 U 1 + U 2 =0 (6:17) 1 2 U 1 + U 2 = U 1 + U 2 =0 (6:18) U 1 ;U 2 U 1 ;U 2 (1) S = U1 =0 (6:19) U 1 1 = 2 = (6:20)

57 62 53 (617) S = (U 1 ) + (U 2 ) 1 V 2 V 2 = 0 (U 2 1 ) 2 0 (U 2 2 ) 2 1 V 1 2 V 2 = h 1 C V C V 2 i (U1 ) 2 < 0 (621) 1 = 2 U 2 = 0U 1 ( ) 52 C P 0 C V = V 2 = >0 C P >C V >0 U = U 1 (S 1 ;V 1 )+U 2 (S 2 ;V 2 ) ; V 1 + V 2 = ; S 1 ;S 2 U = V 1 + V 2 = 0P 1 V 1 0 P 2 V 2 = 0(P 1 0 P 2 )V 1 =0 1 S 2 S 2 U = 1 = 1 V 1 ad 1 P 1 = P 2 (V 1 ) 0 S V 2 ad 2 (V 2 ) 2 S 2 (V1 ) 2 > 0 (624) adiabatic ad 2 (617) (621) i 01 (Ui ) 1 1 (U 1 )+ 1 2 (U 2 )= 1 (U 1 + U 2 )= 1 [(U 1 + U 2 )] = 0

58 54 6 ( ) F > ad = = C V =C P C P >C V 34 p17 > ad > (3-1) n 1 ;n 2 (U 1 ;V 1 ;S 1 ); (U 2 ;V 2 ;S 2 ) S = n 1 S 1 (U 1 ;V 1 )+n 2 S 2 (U 2 ;V 2 ) (625) (1) S = S 1 n 1 + n 1 1 U 1 + n 1P 1 1 V 1 + S 2 n 2 + n 2 2 U 2 + n 2P 2 2 V 2 (626) 6 (n 1 ;U 1 ;V 1 ); (n 2 ;U 2 ;V 2 ) 3 n 2 n 1 1) n 1 + n 2 = n 2 = 0n 1 2) n 1 U 1 + n 2 U 2 = n 2 U 2 = 0n 1 U 1 0 (U 1 0 U 2 )n 1 3) n 1 V 1 + n 2 V 2 = n 2 V 2 = 0n 1 V 1 0 (V 1 0 V 2 )n 1 3 (1) S = n1 U 1 + P P 2 2 n1 V 1 + h S 1 0 U 1 + P 1 V S2 0 U 2 + P 2 V 2 2 i n1 =0 (627) U 1 ;V 1 ;n = 2 ; P 1 = P 2 ; G 1 = G 2 G = U + PV 0 S 55

59 62 55 (3-2) = n 1 ;n 2 (V 1 ;F 1 ); (V 2 ;F 2 ) F = n 1 F 1 (;V 1 )+n 2 F 2 (;V 2 ) (6:28) (1) F = F 1 n 1 0 n 1 P 1 V 1 + F 2 n 2 0 n 2 P 2 V 2 (6:29) 1) n 1 + n 2 = n 2 = 0n 1 2) n 1 V 1 + n 2 V 2 = n 2 V 2 = 0n 1 V 1 0 (V 1 0 V 2 )n 1 n 2 n 1 (1) F = n 1 (P 1 0 P 2 )V 1 +[(F 1 + P 1 V 1 ) 0 (F 2 + P 2 V 2 )]n 1 =0 (6:30) G = F + PV 2 P 1 = P 2 ; G 1 = G 2 (3-3) = P = 1 G 1 ;G 2 G = n 1 G 1 (;P)+n 2 G 2 (;P) (6:31) (1) G = G 1 n 1 + G 2 n 2 =(G 1 0 G 2 )n 1 =0 (6:32) 2 G 1 = G n 2 n 1 P

60 A B 2 n(r;t) u(r;t) S(t) = Z s u(r;t); n(r;t) dv (6:33) s(u; n) 1 V U S(U; V ) u s(u; n) =ns n ; 1 (6:34) n 64

61 64 57 x j q = 0 d (6:35) dx dx udx dx d dt (udx) =j q(x) 0 j q (x + dx) du dt = 0dj q dx = d2 (6:36) dx 2 j q = d 2 dx 2 =0 = c 1x + c 2 (6:37) c 1 ;c r j q = = 0r 1 j q (6:39) ds = 0 1 r1j q (6:40) j S = = 0r 1 j S + j q 1r 1 3 j q = r(1= ) (6:41) (6:42)

62 58 6 (642) (638) j q 1r 1 = r (638) (639) 2 = r 1 2 > = r2 (6:44) u = C + constant = Dr2 u D = =C (6:46) D (644) r 2 =0 (6:47) (Laplace) a; b 1 ; 2 2rj = = c 1 ; 0d=dr = j = c 1 =2r; (r) =0(c 1 =2)logr+c 2 ; (r) = 1 0f(log r0log a)=(log b0log a)g( ) r (6:48) r =grad ; div j r1j =divj z

63 65 59 r2a =rota = (6:51) r1r= r 2 (6:52) r2r= (Lagrange) n F (x 1 ;x 2 ; :::; x n ) r G k (x 1 ;x 2 ; :::; x n )=0 (k =1; 2; ::; r) (6:53) (x 10 ;x 20 ; :::; x n0 ) G k = nx i x i =0 (k =1; 2;:::;r) (6:54) F = i x i =0 (6:55) (655) n fx i g r 1 (654) r fx j g(j = n0r+1;:::;n) (n0r) fx i g (i =1; 2; :::; n0r) F = n0r i + j x i =0 i (n 0 r) fx i g (i =1; 2;:::;n0 r) (656) =0 r f k g (k =1; 2;:::;r) (n 0 r)+r = n rx k F + k G k = + k x i =0 i k=1 i=1 n fx i g (i =1; 2; i + rx k=1 k=1 i =0 (i =1; 2;:::;n) (6:58)

64 60 6 n fx i g f k g x i0 ( 1 ; 2 ; :::; r ) (6:59) r (653) r f k g (659) F n fx i g G = ax + by + cz 0 1=0 F = x 2 + y 2 + z 2 F + G =(2x + a)x +(2y + b)y +(2z + c)z =0 (6:60) x = 0 a 2 ; y = 0b 2 ; z = 0 c 2 (6:61) x 0 = G = 0 (a2 + b 2 + z 2 ) 2 a a 2 + b 2 + c 2 ; y0 = 2 0 1=0 = 0 (6:62) a 2 + b 2 + c 2 b a 2 + b 2 + c 2 ; z0 = c a 2 + b 2 + c 2 (6:63) F F min = 1 a 2 + b 2 + c 2 (6:64) 52 (3-1) ; ; (1) S + (n 1 + n 2 )+(n 1 U 1 + n 2 U 2 )+(n 1 V 1 + n 2 V 2 ) = (S U 1 + V 1 )n 1 +( )n 1 U 1 +( P )n 1 V 1 + (S U 2 + V 2 )n 2 +( )n 2 U 2 +( P )n 2 V 2 =0 (665) =0 0 = 1 1 = = 2 (6:66) 0 = P 1 1 = P 2 2 P 1 = P 2 (6:67) = 0S 1 + U P 1V 1 1 = 0S 2 + U P 2V 2 2 G 1 = G 2 (6:68)

65 { {,,,,,,,, AB,, AB,, 0, ( ) 0,, (;P) @P = 0S = V 61

66 P P = 0 C = 0V G G G S S S V C P C P 4S 4V 4C P 72-2 liquid gas G l ;G g 62 G l (;P)=G g (;P) (7:1) 2 (;P) P given G g G l b = b (P ) P = P s ( )

67 (;P) (;P) - (Clapeyron)- 2 A(;P) A'( + d; P + dp) A G l (;P)=G g (;P) (7:2) A 0 G l ( + d; P + dp )=G g ( + d; P + dp) P G l (;P)0S l d +V l dp = G g (;P)0S g d +V g P = = V A A' (72) (73) (S g 0 S l )d =(V g 0 V l )dp dp d = S g 0 S l V g 0 V l = L b (7:4) b 4V L b = b (S g 0 S l )=H g 0 H l (7:5) 4V = V g 0 V l V g V l 1 18cc 22` 1000 (solid) dp d = S l 0 S s V l 0 V s = L m (7:6) m 4V L m 4V ' 0 V s <V l V s >V l m

68 64 7 (;P) P P 0 0 1) 2) 3) 4) ( )f27316k, g ( )f6744k,2183, 032g/cm 3 g ( )f5436k, g ( )f1548k,501, 041g/cm 3 g 73 Van der Waals) P = R V 0 b 0 a V 2 (7:7)

69 73 65 P + a V 2 (V 0 b) =R (7:8) a; b b b N A v 0 v 0 a (PV 2 + a)(v 0 b) 0 R V 2 =0 (7:9) V (;P) V (i) (ii) (79) 3 P 3 P > 0 <0 R C = 8a 27b ; V C =3b; P C = a (7:11) 27b 2 a; b P C V C R C = 3 8 (= 0:375:::) a; b 023 Ar 0292 He 031 t = = C ; v = V=V C ; p = P=P C p = 8t 3v v 2 (7:12)

70 p t = = C ; p= P=P C ; v= V=V C 2 1 v t =1:15 t =1:00 t =0:85 G l = G g F l + P s V l = F g + P s V g (7:13) P s F g 0 F = 0P s = V g 0 V V =V l ;Vg (7:14) x = n l =(n l + n g ) V V = xv l +(10 x)v g 1 0 x x = V 0 V l (7:15) V g 0 V

71 73 67 F V V l V g P s (V g 0 V l )=0(F g 0 F l )= Z Vg V l P (;V )dv (7:16) P -V P s P P s 0 V V g V l <0 p145 <0

72 68 7 ( V P 0 V F 0 G P 0 V l V g P s

73 S 1 (;P)=S 2 (;P) ; V 1 (;P)=V 2 (;P) (7:17) M = f(;h) ' 0 ( )H ; 0 ( )=C= (7:18) m M M = f(;h + M) (7:19) M M M = ~ f(;h) (7:20) H (@M=@H) H=0 M = 0 (H + M) M = H (7:21) ( )= = C 0 C ; C = C (7:22) = C H =0 M = H =0 - (Weiss)

74 70 7 H =2:0 c c 0 M =0:5 c c :5 c Ṁ H 0 c H M (719) M(H; ) y = f 01 (M) y = M + H f(;h)=tanh CH (7:23) (719) H M H = 1+M log 2C 1 0 M 0 = C(1 0 M 2 ) 0 =0 M 2 =10 C C C H = C = C log 1+M 2 C 1 0 M 0 M (7:25) (H =0) =2M log 1+M (7:26) C 1 0 M

75 H=0 M = 6 C 2( C 0 ) ( C) (7:27) s 1 0 ds = d + dp = C P d 0 V dp dv = d + dp = V d 0 V P (717) - S 1 (;P)=S 2 (;P) V 1 (;P)=V 2 (;P) dp d = 1 V C P 1 0 C P (7:31) dp d = (7:32) (Ehrenfest) (C P 1 0 C P 2 )( )=V( ) = 0 V 3R 0 9R CV C 3V 0 V C 8V 2 0 P = a V 2 U(;V )=U 0( ) 0 a V (7:34) U 0 ( ) C V = = du d = C V 0 (7:35) C P (= C V + V 2 =) dp = V 3R 3V 0 V C d 0 9R[4V3 0 C V C (3V 0 V C ) 2 ] 4V 3 (3V 0 V C ) 2 dv (7:36)

76 72 7 = = 0 P = = 4V 2 (3V 0 V C ) 3[4V 3 0 C V C (3V 0 V C ) 2 ] 4V 2 (3V 0 V C ) 9R[4V 3 0 C V C (3V 0 V C ) 2 ] (7:37) (7:38) [:::::] =4( 0 C )V 3 +(4V 0 V C ) C (V 0 V C ) 2 (7:39) C P = C V 0 ( )+ V C =0 4R V 3 4( 0 C )V 3 +(4V 0 V C ) C (V 0 V C ) 2 (7:40) C P 0 C V ' R t t = 0 C C ) (7:41) < C F = F 0 ( )+R log(3v 0 V C )+ 9R CV C (7:42) S = 0 = S 0 ()+Rlog(3V 0 V C ) V = C (1 + t) V = V C (1 + v) t<0 v =0 v v l ;v g P = 3R C(1 + t) 2V C (1 + 3v=2) 0 9R C 8V C (1 + v) 2 = P C h 1+4t( v + :::) v3 + ::: i (744) P 0 P C P C =4t( v + :::) v3 =4t (7:45) q v = 62 jtj (7:46) S = S 0 ( )+R log 2V C + R 2 log(1 0 9jtj) = ds V d + 9R C V = C V 0 + 9R 2 C 2 ; 4C V = 9R 2 C P (740) (7:48) =4 C V 3(t + 3 C 4 v2 + :::) ' 8 C V 3 Cjtj (7:49) C P 0 C V ' R 2jtj (7:50)

77 lim S 1 (;P 1 )=lims 2 (;P 2 ) (7:51)!0!0 (Nernst) lim S(;P)=0 (7:52)!0 ( - (S; ) P 1 P 2 A( 1 ;P 1 ) B( 1 ;P 2 ) C( 2 ;P 1 ) S B = S( 1 ;P 2 )=S(0;P 2 )+ S C = S( 2 ;P 1 )=S(0;P 1 )+ S C = S B Z 1 0 Z 2 0 C P d (7:53) C P d (7:54) Z 1 S 1 S 2 0 S C 2 A 1 B P = P 1 P = P 2 4S = S(0;P 1 ) 0 S(0;P 2 )= 0 Z C P d 0 2 C P d (7:55) 0

78 74 7 4S >0 1 4S 2 =0 2 =0 4S >0 (1)! 0 ( ) 0 lim!0 C P =0 S = P S lim S = = P S S C P! 0 )= lims +limc P!0!0 C P ( ) / 3 3 C P ( ) / S P = P 1 P = P 2 (2) 0 = (3) 0 lim!0 =0 V S(;V )= = V d ( @ Z @P d 0 = V ( ) V d 0 = ( =0) V @V d 0 = 2 2 d 0

79 8 81 ; V; U; H; S; F; n N (;P) V (;P;n)=nV (;P;1) ; U(;P;n)=nU(;P;1) etc: (8:1) 1 (S; P ) V (S; P; n) 6= nv (S; P; 1) = nv (S=n; P; 1) (8:2) (;P) (;P) G(;P) G(;P;n)=nG(;P;1) (8:3) n dg = 0Sd + VdP+ dn (8:4) ;P (=G(;P;1) ) (8:5) (83) dg = nd + dn (84) 0 Sd + VdP 0 nd =0 (8:6) 1 75

80 (Duhem) F (;V;n) 6= = ;V S;V 82 fn 1 ;n 2 ; :::; n g - G(;P;n 1 ;n 2 ;:::;n ) (8:7) k = k ;P; n 0 k n 0 k k 2 G(;P;n 1 ;n 2 ; :::; n )=G(;P;n 1 ;n 2 ;:::;n ) (8:9) =1 X n k = G G k ;P; n 0 k X k=1 n k k (8:10) n k k 2 p dg = 0Sd + VdP + X k=1 k dn k = X k=1 k dn k + X k=1 n k d k (8:11) f (x 1 ;x 2 ; :::) = p f(x 1 ;x 2 ;:::) X k k = pf (Euler)

81 Sd + VdP 0 X k=1 n k d k =0 (8:12) ( +2) x k = n k Pj n j (8:13) k = k (;P;fx j g) X x k =1 (8:14) k G = X b=1 G (b) (;P;n 1 (b) ;n 2 (b) ;:::;n (b) ) (8:15) 2+ (1) (1) (2) () k = k = ::::: = k (k =1; 2;:::::;) (8:16) ( 0 1) (2) - 0 Sd + VdP0 X k n k (b) d k (b) =0 (b =1; 2; :::::; ) (8:17) (2 + ) 0 ( 0 1) 0 =2+ 0 f = (8:18) =1 3 =1! f =2 (;P) =2! f =1 = (P) P = P s ( ) =3! f =0

82 =2 4 (1) =1 f =3 (;P) (2) =2 f =2 (;P) x 0 = n0 2 n n 0 2 ; x 00 = n00 2 n n 00 2 (8:19) x 0 = x 0 (;P) ; x 00 = x 00 (;P) (8:20) 1 x 00 () x 0 ( ) x 0 x 0 x x G =(n n0 2 )G0 (;P;x 0 )+(n n00 2 )G00 (;P;x 00 ) (8:21) 0 1 = 00 1 (= 1 ) ; 0 2 = 00 2 (= 2 ) (8:22) G 00 0 G 0 x 00 0 x 0 = 2 (;P) 0 1 (;P) (8:25) x 0 (x 00 ;P 0 = 2 (;P)0 1 ;P 00 (8:26) G 0 =(10 x 0 ) 1 (;P)+x 0 2 (;P) (8:23) G 00 =(10 x 00 ) 1 (;P)+x 00 2 (;P) (8:24) G 0 G 00 x 0 x 0 x x 00 1 G 0 G 00

83 85 79 x n 0 x 0 + n 00 x 00 =(n 0 + n 00 )x n 0 (x 0 x 0 )=n 00 (x 00 0 x) (8:27) G = n 0 G 0 (;P;x 0 )+n 00 G 00 (;P;x 00 ) (n 0 + n 00 ) minfg 0 (x); G 00 (x)g (8:28) 85 A,B (;P) (n 1 ;V 1 ) (n 2 ;V 2 ) 56 i) V 1 +V 2 ds = PdV ds = PdV = n irdv V (i =1; 2) (8:29) 4S i = n i R log V 1 + V 2 V i = n i R log n 1 + n 2 n i = 0n i R log x i (8:30) 4S = 0R(n 1 log x 1 + n 2 log x 2 ) (8:31) ii) ds =0 A( ) B d 0 Q = d 0 W =0 du =0 P A A B B A B

84 80 8 i) 4S 4S >0 fn i g x i = n i = P j n j 3 X 4S = 0R n i log x i (8:32) i S = X k n k [ S 0 k 0 R log x k ] ; U = X k n k U 0 k ; H = X k n k H 0 k (8:33) G = H 0 S = X k = X k n k [ H 0 k 0 S0 k + R log x k ] n k [G 0 k + R log x k ] (834) 0 4 k = 0 k + R log x k k G 0 k 0 k (832) 4S >0 P 4S =0 { { 3 4S >0 x i < 1 P 4 G [] x j = n j = n l l n k P log x j =logn j 0 log( n l log x j =@n k = jk =n k 0 1= X X l n j (@ log x j =@n k )=10 n l ( jk =1 if j = k; jk = 0 otherwise) X n j = X n l =10 1=0 j j l

85 H 2 O 2 *) 2H 2 O X i i X i *) X j f i g fn k g n 1 : n 2 : :::::: = 1 : 2 : ::::::: (8:36) n k = 0 k (8:37) (;P) (834) (835) (837) G = X k 0 jx 0 j 0 k (;P)+R log x k nk h X i = 0 k 0 k (;P)+R log x k =0 (838) k [:::::] =0 5 5 [:::::] X k 0 (;P)+R log x k k k = A

86 82 8 (838) X k k log x k = 0 1 R Y k K(;P)=exp h 0 1 R X k k 0 k(;p) (8:39) x k k = K(;P) (8:40) X k k 0 k (;P)i (8:41) (;P) X k x k X k [H 2 ] 2 [O 2 ] [H 2 O] 2 = K(;P) (8:42) - K(;P) (;P) 1) log K(;P)=0 1 log = 0 1 R X k X k k 0 k(;p) 0 k (8:44) 0 k (@0 k=@p ) = V = log K X =(0 k )=P k dg = 0Sd + VdP 0 X a A a d a

87 87 83 (0 P k k ) 4 =1 2+1=3 2 K P P K log P = 0 1 R X 0 k = ) P = 1 X k H 0 R 2 k (8:46) k 51 - X k k H 0 k = Q (8:47) 4 =1 Q >0 Q <0 Q>0 K K (le Chatrier)- (Braun) 87 n 1 ;n 2 x = n 2 n 1 + n 2 = c = n 2 n 1 ( 1) (8:48) c 1+c = c(1 0 c + c2 0 :::::) ( 1) (8:49)

88 84 8 V (;P;n 1 ;n 2 ) = n 1 V c=0 = n 1 [V (;P;1; 0) + c + ;P = n 1 V 1 (;P)+n 2 V 2 (;P) (+::::) (850) (:::::) V 1 (;P)=V(;P;1; 0) V 2 (;P)=(@V=@c) c=0 1m 3 1` U(;P;n 1 ;n 2 )=n 1 U 1 (;P)+n 2 U 2 (;P)+:::; U 2 ;P (8:51) c =0 6 U; V ds = du + PdV = n 1(dU 1 + PdV 1 ) + n 2(dU 2 + PdV 2 ) (8:52) n 1 ;n 2 n 1 (du 1 + PdV 1 ) = n 1 ds 1 du 2 + PdV 2 S = d 0 n 1S 1 n 2 n 2 (8:53) S 2 ds = n 1 ds 1 + n 2 ds 2 (8:54) S(;P;n 1 ;n 2 )=n 1 S 1 (;P)+n 2 S 2 (;P)+C(n 1 ;n 2 ) (8:55) 6 4S = 0R(n 1 log x 1 + n 2 log x 2 )=Rn 1 [(1 + c) log(1 + c) 0 c log c] c =0

89 87 85 (;P) P! 0 C(n 1 ;n 2 )=0R(n 1 log x 1 + n 2 log x 2 ) (8:56) c =0 c G(;P;n 1 ;n 2 )=n 1 G 1 (;P)+n 2 G 2 (;P)+R(n 1 log x 1 + n 2 log x 2 )+::: (8:57) 1 = G 1 (;P)+R log(1 0 x) (8:58) 2 = G 2 (;P)+R log x (8:59) G 1 (;P) G 2 G2 d = V 2 dp 0 U 2 + PV 2 d (8:60) 2 1 (1) G g (;P)=G l (;P)+R log(1 0 x) (8:61) G(;P;1; 0) G l (;P) P G g (;P)=G l (;P) b x 1 b 4 = 0 b log(1 0 x) '0x R log(1 0 x) = G g(;p) 0 G l(;p) G g G l i 4 = R b 2 = 0 H 0 g 0 H 0 l b 2 4 (862) L b x ( L b = H 0 g 0 H 0 l ) (8:63)

90 86 8 (2) (1) 4 = 0 R m 2 L m x (8:64) (3) (Van't Ho) (1) 0 Rx ' i = 4P (V g 0 V l ) (8:65) 4P = 0 R V g 0 V l x (8:66) (4) G 0 (;P 0 )=G 00 (;P 00 ; 1; 0) + R log(1 0 x) (8:67) G 0 G 00 0R x = G(;P 0 ) 0 G(;P 0 + 4P 4P = V 1 4P (868) V 1 5 (= 4P) = R V 1 x (8:69) 4P P 00 P 0 V = n 1 V 1 n 2 =(n 1 + n 2 )x 5V = n 2 R (8:70)

91 87 87 (5) (Henry) G g (;P)=G 2 (;P)+R log x (8:71) P V g = V 2 + log (8:72) V 2 V 2 V log = V g R = 1 P x = c( )P (8:73) P c( ) ( ) 72 n 1 n 2 G g (;P)+R log x = G l (;P) (8:74) x = n 2 n 1 + n 2 = P 2 P (8:75) G g (;P)=G g (;P 0 )+ R log(p=p 0 ) P n 1 n 2 +

92 88 8 G g (;P)+R log x = G g (;P)+R log P 2 P = G g(;p 2 ) (8:76) G g (;P 2 )=G l (;P) (8:77) P P G given P s P >P s (877) G g P 2 x = P 2 =P G l 100% P <P s P 2 >P P P P s P 2 V l G l (;P) ' G l (;P s )+V l (P 0 P s ) P 2 = P s exp[v l (P 0 P s )=R ]; x 2 =(P s =P )exp[v l (P 0 P s )=R ]

93 9 (1) p6 (115) U V U(;V ) P V P U(;V (;P)) P (2) p16 V = C V = C P 0 C V V C V ;C P ; V U(;V ) (3) p17 U U U H F (4) p21 89! P

94 90 9 (5) p27 (41) (334) Q 2 Q 1 = 2 1 (6) p32 (7) p40 (1) V = V (P; @ =(01=V )(@V=@P ) ad =(01=V )(@V=@P) S 0 ad = V 2 =C P C P =C V = = ad (524) > ad (8) p44 S S = s()v @ V F (9) p48 (4) S(;P) G(;P) P P = du d S

95 91 (10) p53 (621) (11) p53 du < 0 du > 0 (12) p53 2V V 1 ; 2 1 (13) p80 (835) (574) G 0 (;P)=G 0 (;P 0 )+R log P=P 0 (P 0 ) x i x i = P i =P (835) k (;P)= 0 k (;P k) p88 05

96 (Ehrenfest) , 86 (Euler) 76 Ostwald , Carnot (Gibbs) 38 - (Duhem) (Curie) 46 - (Weiss) 69 66, Clausius Kelvin , , , , (Joule) 14 17, 40, 48 - (Stefan-Boltzmann) 44 98

97 , , , 73 61, , , , , (homson) 23 (Dalton) 48 12, , (caloric) , 32, , (Nernst)

98 Van der Waals) 64 (Van't Ho) (Helmholtz) 38 3 (Henry) 87 Boyle 10, (Maxwell) (Laplace) (le Chatrier)- (Braun) 83 (LeGendre)

6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1

6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1 6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) (e) Γ (6.2) : Γ B A R (reversible) 6-1 (e) = Clausius 0 = B A: Γ B A: Γ d Q A + d Q (e) B: R d Q + S(A) S(B) (6.3) (e) // 6.2 B A: Γ d Q S(B) S(A) = S (6.4) (e) Γ (6.5)

More information

i 18 2H 2 + O 2 2H 2 + ( ) 3K

i 18 2H 2 + O 2 2H 2 + ( ) 3K i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................

More information

3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3) P = U V S,N,, µ = U N. (4) S,V,, ( ) ds = 1 T du + P T dv µ dn +, (5) T 1 T = P U V,N,, T

3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3) P = U V S,N,, µ = U N. (4) S,V,, ( ) ds = 1 T du + P T dv µ dn +, (5) T 1 T = P U V,N,, T 3 3.1 [ ]< 85, 86 > ( ) ds > 0. (1) dt ds dt =0, S = S max. (2) ( δq 1 = TdS 1 =0) (δw 1 < 0) (du 1 < 0) (δq 2 > 0) (ds = ds 2 = TδQ 2 > 0) 39 3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3)

More information

スライド 1

スライド 1 (version 2011/9/27) 2 1 H i 1 1 2 5 21 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 22 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 23 : : : : : : : : :

More information

現代物理化学 2-1(9)16.ppt

現代物理化学 2-1(9)16.ppt --- S A, G U S S ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r S -- ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r d Q r e = P e = P ΔS d 'Q / e (d'q / e ) --3,e Q W Q (> 0),e e ΔU = Q + W = (Q + Q ) + W = 0

More information

P F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2

P F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2 1 1 2 2 2 1 1 P F ext 1: F ext P F ext (Count Rumford, 1753 1814) 0 100 H 2 O H 2 O 2 F ext F ext N 2 O 2 2 P F S F = P S (1) ( 1 ) F ext x W ext W ext = F ext x (2) F ext P S W ext = P S x (3) S x V V

More information

2012 A, N, Z, Q, R, C

2012 A, N, Z, Q, R, C 2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)

More information

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b) 5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h 0 f(a + h, b) f(a, b) h............................................................... ( ) f(x, y) (a, b) x A (a, b) x

More information

2 p T, Q

2 p T, Q 270 C, 6000 C, 2 p T, Q p: : p = N/ m 2 N/ m 2 Pa : pdv p S F Q 1 g 1 1 g 1 14.5 C 15.5 1 1 cal = 4.1855 J du = Q pdv U ( ) Q pdv 2 : z = f(x, y). z = f(x, y) (x 0, y 0 ) y y = y 0 z = f(x, y 0 ) x x =

More information

September 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1

September 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1 September 25, 2017 1 1.1 1.2 p = nr = 273.15 + t : : K : 1.3 1.3.1 : e.g. 1.3.2 : 1 intensive variable e.g. extensive variable e.g. 1.3.3 Equation of State e.g. p = nr X = A 2 2.1 2.1.1 Quantity of Heat

More information

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ

More information

flMŠÍ−w−î‚b

flMŠÍ−w−î‚b 23 6 30 i 2 1980 2001 1979 K. 1971 ii 1992 iii 1 1 2 5 2.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 5 2.2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8 2.3 : : : : : : : : :

More information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

応力とひずみ.ppt

応力とひずみ.ppt in yukawa@numse.nagoya-u.ac.jp 2 3 4 5 x 2 6 Continuum) 7 8 9 F F 10 F L L F L 1 L F L F L F 11 F L F F L F L L L 1 L 2 12 F L F! A A! S! = F S 13 F L L F F n = F " cos# F t = F " sin# S $ = S cos# S S

More information

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x ( II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )

More information

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................ 5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h f(a + h, b) f(a, b) h........................................................... ( ) f(x, y) (a, b) x A (a, b) x (a, b)

More information

i

i i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy

More information

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2 3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1

More information

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0, .1.1 Y K L Y = K 1 3 L 3 L K K (K + ) 1 1 3 L 3 K 3 L 3 K 0 (K + K) 1 3 L 3 K 1 3 L 3 lim K 0 K = L (K + K) 1 3 K 1 3 3 lim K 0 K = 1 3 K 3 L 3 z = f(x, y) x y z x-y-z.1 z = e x +xy y 3 x-y ( ) z 0 f(x,

More information

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

More information

入試の軌跡

入試の軌跡 4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

II 2 II

II 2 II II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................

More information

( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n

More information

ii

ii ii iii 1 1 1.1..................................... 1 1.2................................... 3 1.3........................... 4 2 9 2.1.................................. 9 2.2...............................

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi) 0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

More information

30

30 3 ............................................2 2...........................................2....................................2.2...................................2.3..............................

More information

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x [ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),

More information

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 - M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................

More information

2011de.dvi

2011de.dvi 211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37

More information

( )

( ) 18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................

More information

y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =

y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a = [ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =

More information

Untitled

Untitled II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j

More information

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2 No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

More information

r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B

r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B 1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n

More information

5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E

5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 2, S 1 N 1 = S 2 N 2 2 (chemical potential) µ S N

More information

1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載

1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載 1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります

More information

genron-3

genron-3 " ( K p( pasals! ( kg / m 3 " ( K! v M V! M / V v V / M! 3 ( kg / m v ( v "! v p v # v v pd v ( J / kg p ( $ 3! % S $ ( pv" 3 ( ( 5 pv" pv R" p R!" R " ( K ( 6 ( 7 " pv pv % p % w ' p% S & $ p% v ( J /

More information

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx

More information

70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain) (grain boundary) 5. 1

70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain) (grain boundary) 5. 1 5 0 1 2 3 (Carnot) (Clausius) 2 5. 1 ( ) ( ) ( ) ( ) 5. 1. 1 (system) 1) 70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain)

More information

2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a

More information

II 2014 2 (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1

II 2014 2 (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1 II 2014 1 1 I 1.1 72 r 2 72 8 72/8 = 9 9 2 a 0 1 a 1 a 1 = a 0 (1+r/100) 2 a 2 a 2 = a 1 (1 + r/100) = a 0 (1 + r/100) 2 n a n = a 0 (1 + r/100) n a n a 0 2 n a 0 (1 + r/100) n = 2a 0 (1 + r/100) n = 2

More information

2013 25 9 i 1 1 1.1................................... 1 1.2........................... 2 1.3..................................... 3 1.4..................................... 4 2 6 2.1.................................

More information

6. Euler x

6. Euler x ...............................................................................3......................................... 4.4................................... 5.5......................................

More information

// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du

// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du 2 2.1 1 [ 1 ]< 33, 34 > 1 (the first law of thermodynamics) U du = δw + δq (1) (internal energy)u (work)w δw rev = PdV (2) P (heat)q 1 1. U ( U ) 2. 1 (perpetuum mobile) 3. du 21 // //( ) (Helmholtz, Hermann

More information

1 B () Ver 2014 0 2014/10 2015/1 http://www-cr.scphys.kyoto-u.ac.jp/member/tsuru/lecture/... 1. ( ) 2. 3. 3 1 7 1.1..................................................... 7 1.2.............................................

More information

0 (Preliminary) T S pv

0 (Preliminary) T S pv 0 (Preliminary) 4 0.1.................. 4 0.1.1 S p................... 4 0.1.2......................... 5 0.2............................. 8 0.2.1 p S................ 11 0.3..............................

More information

all.dvi

all.dvi I 1 Density Matrix 1.1 ( (Observable) Ô :ensemble ensemble average) Ô en =Tr ˆρ en Ô ˆρ en Tr  n, n =, 1,, Tr  = n n  n Tr  I w j j ( j =, 1,, ) ˆρ en j w j j ˆρ en = j w j j j Ô en = j w j j Ô j emsemble

More information

mugensho.dvi

mugensho.dvi 1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim

More information

熊本県数学問題正解

熊本県数学問題正解 00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (

More information

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

More information

Acrobat Distiller, Job 128

Acrobat Distiller, Job 128 (2 ) 2 < > ( ) f x (x, y) 2x 3+y f y (x, y) x 2y +2 f(3, 2) f x (3, 2) 5 f y (3, 2) L y 2 z 5x 5 ` x 3 z y 2 2 2 < > (2 ) f(, 2) 7 f x (x, y) 2x y f x (, 2),f y (x, y) x +4y,f y (, 2) 7 z (x ) + 7(y 2)

More information

untitled

untitled 20010916 22;1017;23;20020108;15;20; 1 N = {1, 2, } Z + = {0, 1, 2, } Z = {0, ±1, ±2, } Q = { p p Z, q N} R = { lim a q n n a n Q, n N; sup a n < } R + = {x R x 0} n = {a + b 1 a, b R} u, v 1 R 2 2 R 3

More information

st.dvi

st.dvi 9 3 5................................... 5............................. 5....................................... 5.................................. 7.........................................................................

More information

2009 June 8 toki/thermodynamics.pdf ) 1

2009 June 8   toki/thermodynamics.pdf ) 1 2009 June 8 http://www.rcnp.osaka-u.ac.jp/ toki/thermodynamics.pdf 1 6 10 23 ) 1 H download 2 http://www.rcnp.osaka-u.ac.jp/ toki/thermodynamics.pdf 2 2.1 [1] [2] [3] Q = mc (1) C gr Q C = 1cal/gr deg

More information

http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 2 N(ε 1 ) N(ε 2 ) ε 1 ε 2 α ε ε 2 1 n N(ɛ) N ɛ ɛ- (1.1.3) n > N(ɛ) a n α < ɛ n N(ɛ) a n

http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 2 N(ε 1 ) N(ε 2 ) ε 1 ε 2 α ε ε 2 1 n N(ɛ) N ɛ ɛ- (1.1.3) n > N(ɛ) a n α < ɛ n N(ɛ) a n http://www2.math.kyushu-u.ac.jp/~hara/lectures/lectures-j.html 1 1 1.1 ɛ-n 1 ɛ-n lim n a n = α n a n α 2 lim a n = 1 n a k n n k=1 1.1.7 ɛ-n 1.1.1 a n α a n n α lim n a n = α ɛ N(ɛ) n > N(ɛ) a n α < ɛ

More information

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D

1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA  appointment Cafe D 1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00

More information

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx 4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan

More information

2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =

More information

0 = m 2p 1 p = 1/2 p y = 1 m = 1 2 d ( + 1)2 d ( + 1) 2 = d d ( + 1)2 = = 2( + 1) 2 g() 2 f() f() = [g()] 2 = g()g() f f () = [g()g()]

0 = m 2p 1 p = 1/2 p y = 1 m = 1 2 d ( + 1)2 d ( + 1) 2 = d d ( + 1)2 = = 2( + 1) 2 g() 2 f() f() = [g()] 2 = g()g() f f () = [g()g()] 8. 2 1 2 1 2 ma,y u(, y) s.t. p + p y y = m u y y p p y y m u(, y) = y p + p y y = m y ( ) 1 y = (m p ) p y = m p y p p y 2 0 m/p U U() = m p y p p y 2 2 du() d = m p y 2p p y 1 0 = m 2p 1 p = 1/2 p y

More information

A

A A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................

More information

1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1

1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1 1/5 ( ) Taylor ( 7.1) (x, y) f(x, y) f(x, y) x + y, xy, e x y,... 1 R {(x, y) x, y R} f(x, y) x y,xy e y log x,... R {(x, y, z) (x, y),z f(x, y)} R 3 z 1 (x + y ) z ax + by + c x 1 z ax + by + c y x +

More information

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

More information

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

More information

C:/KENAR/0p1.dvi

C:/KENAR/0p1.dvi 2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x

More information

Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math))

Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math)) Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math)) 2001 1 e-mail:s00x0427@ip.media.kyoto-u.ac.jp 1 1 Van der Pol 1 1 2 2 Bergers 2 KdV 2 1 5 1.1........................................

More information

v er.1/ c /(21)

v er.1/ c /(21) 12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,

More information

I 1

I 1 I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg

More information

0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9

0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9 1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),

More information

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

More information

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C 0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

More information

K E N Z OU

K E N Z OU K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................

More information

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10% 1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n

More information

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n ( 3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc

More information

DVIOUT

DVIOUT A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)

More information

0 (Preliminary) F G T S pv (1)

0 (Preliminary) F G T S pv (1) 0 (Preliminary) 4 0.1 F G.............. 4 0.1.1 S p............... 4 0.1.2......................... 6 0.2............................. 7 0.2.1 (1) p, S............................. 12 0.2.2 (2) 13 0.3

More information

20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

現代物理化学 1-1(4)16.ppt

現代物理化学 1-1(4)16.ppt (pdf) pdf pdf http://www1.doshisha.ac.jp/~bukka/lecture/index.html http://www.doshisha.ac.jp/ Duet -1-1-1 2-a. 1-1-2 EU E = K E + P E + U ΔE K E = 0P E ΔE = ΔU U U = εn ΔU ΔU = Q + W, du = d 'Q + d 'W

More information

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1

More information

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq 49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

More information

21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

More information

pdf

pdf http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

More information

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g( 06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,

More information

(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n

(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n . 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n

More information

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f ,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)

More information

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2 II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

More information

Microsoft Word - ●ipho-text3目次

Microsoft Word - ●ipho-text3目次 国際物理オリンピック 研修用テキスト Ⅲ 熱物理 相対論 量子力学 特定非営利活動法人物理オリンピック日本委員会 1 1.1 1 1. 1.3 3 1.4 4 1.5 6 1.6 7 1.7 9 11.1 11. 0.3 1 6 3.1 6 3. -9 3.3 - -- 31 3.4 --33 39 4.1 39 4. 40 4.3 4 4.4 44 4.5 47 5 5.1 5 5. 5 5.3

More information

notekiso1_09.dvi

notekiso1_09.dvi 39 3 3.1 2 Ax 1,y 1 Bx 2,y 2 x y fx, y z fx, y x 1,y 1, 0 x 1,y 1,fx 1,y 1 x 2,y 2, 0 x 2,y 2,fx 2,y 2 A s I fx, yds lim fx i,y i Δs. 3.1.1 Δs 0 x i,y i N Δs 1 I lim Δx 2 +Δy 2 0 x 1 fx i,y i Δx i 2 +Δy

More information

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II ( ) (7/31) II (  [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier

More information

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V

More information

( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)

( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r) ( : December 27, 215 CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x f (x y f(x x ϕ(r (gradient ϕ(r (gradϕ(r ( ϕ(r r ϕ r xi + yj + zk ϕ(r ϕ(r x i + ϕ(r y j + ϕ(r z k (1.1 ϕ(r ϕ(r i

More information

B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t), y(t), z(t)), a t b.

B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t), y(t), z(t)), a t b. 2009 7 9 1 2 2 2 3 6 4 9 5 14 6 18 7 23 8 25 9 26 10 29 11 32 12 35 A 37 1 B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t),

More information

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33

More information