- #1
Jenkz
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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]