KTGHard
Question
One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by PVn = constant. We can conclude that (where T = temperature, W = work done by gas, V = volume and U = internal energy ).
Options
A.TA = TB = TC
B.VA < VB , PB < PC
C.WAB < WBC
D.UA < UB
Solution
(D)
Process AB is isobaric [V α T]
TB > TA ∴ UB > UA
WBC < WAB (Area under P-V curve)
Process AB is isobaric [V α T]
TB > TA ∴ UB > UA
WBC < WAB (Area under P-V curve)
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