Physical Chemistry Third Edition

2.4 Calculation of Amounts of Heat and Energy Changes 67

b.Calculateqandwfor the reversible process corresponding to the following path. Step 1: The

system is heated from 298.15 K to 373.15 K at a constant volume of 2.000 L; step 2: It is then

expanded isothermally to a volume of 20.000 L.

c.Comment on the difference between theqandwvalues for parts a and b. What is the value

of∆Ufor the process of part b?

Reversible Adiabatic Processes

Anadiabatic processis one in which no heat is transferred to or from a closed system,

so thatdqis equal to zero for every infinitesimal step of the process:

dUdq+dwdw (adiabatic process) (2.4-16)

Consider a reversible adiabatic process of an ideal gas.

dUCVdT (closed ideal gas, reversible process) (2.4-17)

dw−PdV−

nRT

V

dV (closed ideal gas, reversible process) (2.4-18)

Becausedq0 for an adiabatic process, we equatedUanddw:

CVdT−

nRT

V

dV (closed ideal gas, reversible adiabatic process) (2.4-19)

This is adifferential equationthat can be solved to giveTas a function ofVif the

dependence ofCVonTandVis known.

We first assume thatCVis constant. We can solve Eq. (2.4-19) by separation of

variables. We divide byTto separate the variables (remove anyVdependence from

the left-hand side and anyTdependence from the right-hand side):

CV

T

dT−

nR

V

dV (2.4-20)

Because each integrand contains only one variable we can integrate Eq. (2.4-20) from

the initial state, denoted byV 1 andT 1 , to the final state, denoted byV 2 andT 2. A definite

integration gives

CVln

(

T 2

T 1

)

−nRln

(

V 2

V 1

)

We divide byCV and take the exponential (antilogarithm) of both sides of this

equation:

T 2

T 1



(

V 1

V 2

)nR/CV



(

V 1

V 2

)R/CV, m

(reversible adiabatic process,

ideal gas,CVconstant)

(2.4-21a)