1000 Solved Problems in Modern Physics

(Romina) #1

278 4 Thermodynamics and Statistical Physics


4.39 By Maxwell’s first equation
(
∂S
∂V


)

T

=

(

∂P

∂T

)

V

(1)

dS=

dU+PdV
T

(2)

using (2) in (1)
(
∂U
∂V

)

T

=T

(

∂P

∂T

)

−P

For perfect gases,

P=

RT

( V

∂U

∂V

)

T

=

RT

V

−P= 0

Thus, temperature remaining constant, the internal energy of an ideal gas
is independent of the volume.

4.40

dP
dT

=

L

T(ν 2 −ν 1 )

dT=

T

L

(ν 2 −ν 1 )dP

=

373(1677−1)(2× 106 )

546 × 4. 2 × 107

= 55. 1 ◦C

4.41 ν 1 =1cm^3 ;v 2 =


1

0. 091

= 10 .981 cm^3

dP=

LdT
T(ν 2 −ν 1 )

=

80 × 4. 2 × 107 × 1

(− 1 +273)(10. 981 − 1 .0)

= 1. 238 ×

106 dynes
cm^2

= 1 .24 atm

P 2 =P 1 +dP= 1. 0 + 1. 24 = 2 .24 atm

4.42 ν 1 =


1

ρ 1

=

1

1. 145

= 0 .873 cm^3 /g

ν 2 =

1

ρ 2

=

1

0. 981

= 1 .019 cm^3 /g

dT=

T(ν 2 −ν 1 )dP
L

=

(80+273)(1. 019 − 0 .873)(1. 0 × 106 )

35. 5 × 4. 2 × 107

= 0. 0346 ◦C
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