Physical Chemistry , 1st ed.

Recall that, for an adiabatic process,

dUdw

because heat is exactly zero. From equation 2.21, we also know that

dUCVdT+ 

U

V



T

dVCVdT for an ideal gas

where the final equality recognizes that the partial derivative ( U/ V)Tequals

zero for an ideal gas. Therefore, for an infinitesimal adiabatic process,

dwCVdT

Integrating for the overall adiabatic process,

w

Ti

Tf

CVdT (2.41)

For a constant heat capacity,

wCVT (2.42)

For anything other than 1 mole, we must use the molar heat capacity,CV:

wnCVT (2.43)

If the heat capacity is not constant over the temperature range, equation 2.41

must be used with the proper expression for CVto calculate the work of the

change.

Example 2.12

Consider 1 mole of an ideal gas at an initial pressure of 1.00 atm and initial

temperature of 273.15 K. Assume it expands adiabatically against a pressure

of 0.435 atm until its volume doubles. Calculate the work, the final temper-

ature, and the Uof the process.

Solution

The volume change of the process must be determined first. From the initial

conditions, we can calculate the initial volume, and then its change:

(1.00 atm)Vi(1 mol)0.08205 

m

L

o

a

l

t



m

K

(273.15 K)

Vi22.4 L

If the volume is doubled during the process, then the final volume is 44.8 L,

and the change in volume is 44.8 L 22.4 L 22.4 L.

The work performed is calculated simply by

wpextV

w(0.435 atm)(22.4 L)

1

L

0



1

a

.3

tm

2J

987 J

Because q0,Uw, so that

U987 J

The final temperature can be calculated using equation 2.43, recognizing that

for an ideal gas the heat capacity at constant volume is ^32 R, or 12.47 J/molK.

Therefore,

48 CHAPTER 2 The First Law of Thermodynamics