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:)