0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# One mole of a vander Waal's gas obeying the equation(p+aV2)(Vâˆ’b)=RTundergoes the quasi-static cyclic process which is shown in the p-V diagram. The net heat absorbed by the gas in this process is:

A
12(p1p2)(V1V2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
12(p1+p2)(V1V2)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
12(p1+aV21p2aV22)(V1V2)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
12(p1+aV21+p2+aV22)(V1V2)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A 12(p1−p2)(V1−V2)Workdone during the complete cycle is equal to the area enclosed by the P-V graph ∴ W=12×(OB)×(OA)=12(p1−p2)(V1−V2)Also the change in internal energy in a complete cycle is Zero i.e ΔU=0Using first law of thermodynamics : Q=ΔU+W∴ Q=0+W=WThus the heat absorbed by the gas in a complete cycle Q=W=12(p1−p2)(V1−V2)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Liquids in Liquids and Raoult's Law
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program