A cyclic process for 1 mole of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively are

Options

A. `0, RT_(1) ln((V_(1))/(V_(2))), R(T_(1) - T_(2))`
B. `R, R(T_(1) - T_(2)), RT_(1) ln ((V_(1))/(V_(2)))`
C. `0, RT_(2) ln ((V_(1))/(V_(2))), R(T_(1) - T_(2))`
D. `0, RT_(2) ln ((V_(2))/(V_(1))) , R(T_(1) - T_(2))`

Answer

D

Explanation

During AB, process is isochoric.
So `Delta V = 0 :. W = 0`
During BC, process is isothermal. So
`Delta T = 0`
`:. W = RT_(2) ln(V_(2))/(V_(1))`
During CA, process is isobaric. So, pressure is constant.
`:. W = P(V_(1) - V_(2))`
But `PV_(1) = RT_(1)` and `PV_(2) = RT_(2) :. W = R(T_(1) - T_(2))`