(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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)

2. Relevant equations

dU= dQ + dW

dW= -p dV

V1= V0/2

3. 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]

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# Homework Help: Adiabatic process in a monatomic ideal gas

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