calculate the work done when one litre of monoatomic perfect gas at ntp is compressed adiabatically to half of its volume use gamma equal to 1.67
Answer (1)
6th Nov, 2020
Hello Aakash!
I am explaining the solution to the question you asked below:-
Work done is an adiabatic process is:-
W =[1/(1-gamma)](P2V2- P1V1)
Here, P1= 10^5 N/m^2 and V1= 6×10^-3 m^3
Given, Cv=3R/2
P2 = P1(V1/V2)^gamma ; V2 = 2×10^-3 m^3
Cp=5R/2 (Cp-Cv=R)
T=Cp/Cv = 1.67
P2= 10^5(6/2)^(1.67) = 10^5×(3)^(1.67) =6.26×10^5 N/m^2
W = [1/(1-1.67)](6.26×10^5×2×10^-3 - 10^5×6×10^-3)
=1/-0.67(1252-600) = -652/0.67 = -973.1 J
The work done is negative as the gas is compressed .
Hope your confusion is clear!
Happy learning:)
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