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
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:)