Physical Chemistry Third Edition

44 2 Work, Heat, and Energy: The First Law of Thermodynamics

EXAMPLE 2.2

a.Calculate the work done on a closed system consisting of 50.00 g of argon, assumed ideal,

when it expands isothermally and reversibly from a volume of 5.000 L to a volume of

10.00 L at a temperature of 298.15 K.

b.Calculate the integral ofdPfor the same process.

Solution

a. w−(50.00 g)

(

1 mol

39 .938 g

)

(8.3145 J K−^1 mol−^1 )(298.15 K) ln

(

10 .00 L

5 .000 L

)

−2151 J

The negative sign indicates that work is done on the surroundings by the system.

b.For a sample of ideal gas with fixednwe have a two-term expression fordP:

dP

(

∂P

∂T

)

V,n

dT+

(

∂P

∂V

)

T,n

dV

nR

V

dT−

nRT

V^2

dV

The change inPfor the finite process is given by integration:

∆P

∫

c

dP

∫T 2

T 1

nR

V

dT+

∫V 2

V 1

nRT

V^2

dV

In order to carry out a line integral like this we must know howVdepends onTfor

the first term and must know howTdepends onVfor the second term. In this case the

process is isothermal (Tis constant). BecausedT0 for each infinitesimal step of the

process, the first term vanishes. In the second term the factornRTcan be factored out of

the integral:

∆P−nRT

∫V 2

V 1

1

V^2

dVnRT

(

1

V 2

−

1

V 1

)

(50.00 g)

(

1 mol

39 .938 g

)

(8.3145 J K−^1 mol−^1 )

×(298.15 K)

(

1

0 .01000 m^3

−

1

0 .005000 m^3

)

− 3. 103 × 105 Jm−^3 − 3. 103 × 105 Pa− 3 .062 atm

Reversible Work Done on a Nonideal Gas

For any nonideal gas equation of state, the expression for the work done in an isothermal

reversible volume change can be obtained by integration.

EXAMPLE 2.3

Obtain the formula for the work done on a sample of gas during an isothermal reversible

volume change if it is represented by the truncated virial equation of state:

PVm

RT

 1 +

B 2

Vm

(2.1-14)