文档介绍：Solutions to Statistical Mechanics Problems
April 30, 2001
1. 1 cal. = ×107 ergs = J. Hence, 639 kcal = 639 × kJ = 2675 kJ.
3
2. For an ideal gas where pV = NkBT and E = 2 NkBT :
!
∂E 3
CV = = NkB
∂T 2
! ! V !
∂Q ∂E ∂V
C p
p = ∂T = ∂T + ∂T
p p p
3
= NkB + NkB
2
Cp − CV = NkB
For an adiabatic QSP process involving an ideal gas, dQ =0and
from E = CV TanddE+ pdV =0
CV dT +(Cp − CV )TdV/V =0
dT/T +(γ− 1)dV/V =0
V γ−1T = constant
or pV γ= constant
3. For adiabatic expansion into vacuum, E is a constant. Hence
2an
T2 = T1 + {1/V2 − 1/V1}
3R
4. From the definition of latent heat:
∆E = L − p∆V
= × 104 − × × 105−2
= × 104J
5. (a) Work done is 1 atm cm3 = × 105 × 10−6J = J.
(b) Work done is 73 × 10−7 J.
Thus in considering macroscopic volume changes of gases, the change in surface energy can be
neglected.
6. The correct statements are:
(a) The entropy change in a reversible cyclic process is zero.
R
(b) The entropy change in a reversible QSP process is dQ/T .
7. Rate of production of heat is I2 , dQ = I2Rdt and dS = I2Rdt/T .
8. (a) In p−V diagram, the adiabats obeying pV γ= K = constant are steeper than the isothermals
where pV = NkBT and are thus curves 2 → 3and4→ 1. Use the equations of state connecting
pi,Vi with each other.
γγ
p2V2 =