One mole of a monoatomic ideal gas is taken along two cyclic processes, $E \to F \to G \to E$ and $E \to F \to H \to E$ as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.
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-II
P $G \to E$ 1 $160 P_{0} V_{0} l n 2$
Q $G \to H$ 2 $36 P_{0} V_{0}$
R $F \to H$ 3 $24 P_{0} V_{0}$
S $F \to G$ 4 $31 P_{0} V_{0}$

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P- 4, Q - 3, R – 1, S - 2

P – 3, Q – 1, R – 2, S - 4

P- 4, Q – 3, R – 2, S - 1

P – 1, Q – 3, R – 2, S - 4

answer is A.

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Detailed Solution

$y_{o n o n u} = \frac{1}{f} = 1 + \frac{2}{3} = \frac{5}{3}$
Isothermal    $F \to G$
$32 p_{0} V_{0} = p_{0} \times V_{G}$    $V_{G} = 32 V_{0}$ $F \to H$ adiabatic
$32 p_{0} {v_{0}}^{5 / 3} = p . {V_{H}}^{5 / 3}$
$V_{H} = 8 V_{0} p . G \to E (I s o b a r i c) △ w = P_{0} (32 V_{0} - V_{0}) = 31 p_{0} V_{0} P \to o f$
$Q .$ workdone in isobaric process $G \to H$
$△ w = P_{0} (32 V_{0} - 8 V_{0}) = 24 P_{0} V_{0}$
$R .$ wokd done in adiabatic process
   $△ w = \frac{32 p_{0} V_{0} - p_{0} 8 V_{0}}{5 / 3 - 1} = 36 D_{0} V_{0}$
$R \to 2$