Adiabatic process in a monatomic ideal gas

[Jenkz]

Homework Statement

In a quasistaic adiabatic process in a monatomic ideal gas PV^5/3 = constant [DO
NOT PROVE]. A monatomic ideal gas initially has a pressure of P0 and a volume of
V0. It undergoes a quasistatic adiabatic compression to half its initial volume. Show
that the work done on the gas is

W = 3/2 P0V0 ( 2^(2/3) - 1)

Homework Equations

dU= dQ + dW
dW= -p dV
V1= V0/2

The Attempt at a Solution

dU= dW as adiabatic procees means dQ=0

dU= 3/2 NKbT = -p dV

And I don't know what to do next.

3/2 NKbT= P0Vo[1-V1/V0]

 

[G01]

You have given the definition of the work done on the gas:

W=\int P dV

HINT: Now if you were given P as a function of V you could calculate the integral explicitly.

 

[Jenkz]

I'm sorry, I don't quite understand what I should do.

 

[Andrew Mason]

Use PV^{5/3}=\text{constant} to find the final P. Then find the temperature from the Pf and Vf. Use that to determine the change in internal energy, which as you have stated must equal the work done by the gas.

AM

 

[Jenkz]

Ok, thank you!