The 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 is

Options

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

Answer

C

Explanation

During AB, process is isochoric
`therefore DeltaV=0`
`thereforeW=0`
During BC, process is
isothermal
`therefore DeltaT=0`
`therefore W=RT_(2)ln(V_(2)/(V_(1)))`

During CA, process is isobarci. So pressure is constant
`therefore W=P(V_(1)-V_(2))`
But `PV_(1)=rT_(1) therefore P=(RT_(1))/(V_(1))=(RT_(2))/(V_(2))`
`therefore W=(RT_(1))/(V_(1))(V_(1)-V__(2))=(RT_(2))/(V_(2))(V_(1)-V_(2))`