One mole of an ideal gas having initial volume V. pressure 2P and temperature T undergoes a cyclic process ABCDA as shown below

The net work done in the complete cycle is

Options

A. Zero
B. `1/2 R T In 2`
C. RT In 2
D. `3/2 R T In 2`

Answer

C

Explanation

`Delta W_(AB)=P(V_2 -V_1)=n R(T_2-T_1)` =RT [P= constant]
`Delta W_(CD)=P(V_1-V_2)= n R(T_1-T_2)`= -RT
`W_(BC)= n R T In((V_2)/(V_1))` [isothermal process]
= 2RT In 2
`W_(DA)=n R T In ((V_1)/(V_2))= n R T In ((P_1)/(P_2))`
= R T In `(1/2)` = -R T In 2
`W_("net") = W_(AB) + W_(CD)+W_(BC) + W_(DA)` = RT In 2