Question

A monatomic ideal gas has an initial temperature of 381 K. This gas expands and does the same amount of work whether the expansion is adiabatic or isothermal. When the expansion is adiabatic, the final temperature of the gas is 290 K. What is the ratio of the final to the initial volume when the expansion is isothermal?

Answer 1

here,

Assuming both processes have the same initial values P0, V0

Final values have suffix a for adiabatic, i for isothermal
Given T0 and Ta
Monatomic gamma = 5/3

Adiabatic:

Va/V0 = 1/(Ta/T0)^(1/(gamma -1))

adiabatic constant AC =

P0*V0^gamma

work = AC(Va^(1-gamma )-V0^(1-gamma ))/(1-gamma )

= P0*V0^gamma *(Va^(1-gamma )-V0^(1-gamma ))/(1-gamma )

Isothermal:

work = P0*V0*ln(Vi/V0)

Equating adiabatic and isothermal work,

ln(Vi/V0) = work/(P0V0) = (P0*V0^gamma *(Va^(1-gamma ) - V0^(1-gamma ))/(1-gamma )) / (P0V0)
= V0^(gamma -1)*(Va^(1-gamma )-V0^(1-gamma ))/(1-gamma )

From adiabatic volume ratio, Va = V0/(Ta/T0)^(1/(gamma -1)) = V0*(Ta/T0)^(1/(1-gamma ))

Then Va^(1-gamma ) = V0^(1-gamma )*Ta/T0

Ta = 290 K and Ti = 381 K,

ln(Vi/V0) = V0^(gamma -1)*(V0^(1-gamma )*Ta/T0 - V0^(1-gamma ))/(1-gamma )

ln(Vi/V0) = (Ta/T0-1)/(1-gamma )

vi/v0 = e^((290/381 - 1)/(1 - 5/3))

Vi/V0 = 1.43

the ratio of the final to the initial volume when the expansion is isothermal is 1.43:1