Physical Chemistry Third Edition

2.4 Calculation of Amounts of Heat and Energy Changes 69

EXAMPLE2.18

Show that for a reversible adiabatic process the van der Waals gas obeys

CVdT−

nRT

V−nb

dV (2.4-23)

Solution

Consider a system containing 1.000 mol of gas. Using Eq. (2.4-22) we can write for a closed

system

dU

(

∂U

∂T

)

V,n

dT+

(

∂U

∂V

)

T,n

dVCVdT+

a

Vm^2

dV

For a reversible process

dw−PdV−

(

RT

Vm−b

−

a

Vm^2

)

dV

For an adiabatic process

dUCVdT+

a

Vm^2

dVdw−

(

RT

Vm−b

−

a

Vm^2

)

dV

so that when terms are canceled

CVdT−

RT

Vm−b

dV

Exercise 2.15

Show that for a reversible adiabatic process in a van der Waals gas with constantCV,m,

(

T 2

T 1

)



(

V 1 −nb

V 2 −nb

)nR/CV

(2.4-24)

EXAMPLE2.19

Find the final temperature for the process of Example 2.17, using Eq. (2.4-24) instead of

Eq. (2.4-21), but still assuming thatCV 3 nR/2.

Solution

From Table A.3, the van der Waals parameterbis equal to 3. 219 × 10 −^5 m^3 mol−^1 for

argon.

T 2 (373.15 K)

(

5. 000 × 10 −^3 m^3 mol−^1 − 3. 22 × 10 −^5 m^3 mol−^1

20. 00 × 10 −^3 m^3 mol−^1 − 3. 22 × 10 −^5 m^3 mol−^1

) 2 / 3

 147 .6K

This value differs from the ideal value by 0.5 K.