Problem 1.12 : Calculate the work done when one litre of a
monoatomic perfect gas at NTP is compressed adiabatically to half its
volume. y = 1.67
Plan
Calin
Given
1
13
Answers
Answered by
1
Answer:
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Explanation:
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:)
Answered by
1
Answer:
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
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