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One mole of a monatomic ideal gas is taken along two cyclic processes $E \rightarrow F \rightarrow G \rightarrow E$ and $E \rightarrow F \rightarrow H \rightarrow$ E as shown in the $PV$ diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. $Image$
Match the paths in List $I$ with the magnitudes of the work done in List $II$ and select the correct answer using the codes given below the lists.
| List $I$ | List $I$ |
| $P.$ $\quad G \rightarrow E$ | $1.$ $\quad 160 P_0 V_0 \ln 2$ |
| $Q.$ $\quad G \rightarrow H$ | $2.$ $\quad 36 P _0 V _0$ |
| $R.$ $\quad F \rightarrow H$ | $3.$ $\quad 24 P _0 V _0$ |
| $S.$ $\quad F \rightarrow G$ | $4.$ $\quad 31 P_0 V_0$ |
Codes: $ \quad \quad P \quad Q \quad R \quad S $
$\quad 4 \quad 3 \quad 2 \quad 1 $
$\quad 4 \quad 3 \quad 1 \quad 2 $
$\quad 3 \quad 1 \quad 2 \quad 4 $
$\quad 1 \quad 3 \quad 2 \quad 4 $
Solution
In $F \rightarrow G$ work done in isothermal proces is $n R T \ln \left(\frac{V_f}{V_i}\right)=32 P_0 V_0 \ln \left(\frac{32 V_0}{V_0}\right)$
$=32 P_0 V_0 \ln 2^5=160 P_0 V_0 \ln 2 $
$\ln G \rightarrow E, \Delta W=P_0 \Delta V=P_0\left(31 V_0\right)=31 P_0 V_0$
In $G \rightarrow H$ work done is less than $31 P_0 V_0$ i.e., $24 P_0 V_0$
In $F \rightarrow H$ work done is $36 P _0 V _0$
