One mole of a monatomic ideal gas is taken along two cyclic processes and as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. [2013]

Question

One mole of a monatomic 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. [2013]

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.	$160 P_{0} V_{0} \ln 2$
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.	$31 P_{0} V_{0}$

Codes :

Smart Achievers · Accepted Answer

(1)

$W_{G E} = P_{0} (V_{0} - 32 V_{0}) = - 31 P_{0} V_{0}$

$W_{G H} = P_{0} (8 V_{0} - 32 V_{0}) = - 24 P_{0} V_{0}$

${(W_{F H})}_{adiabatic} = \frac{P_{f} V_{f} - P_{i} V_{i}}{1 - γ} = \frac{P_{0} (8 V_{0}) - 32 P_{0} (V_{0})}{1 - \frac{5}{3}} = 36 P_{0} V_{0}$

${(W_{F G})}_{isothermal} = n R T \ln (\frac{V_{f}}{V_{i}}) = P_{0} V_{0} \ln (\frac{V_{f}}{V_{i}})$

$= 1 (32 P_{0} V_{0}) \log_{e} (\frac{32 V_{0}}{V_{0}})$

$= 32 P_{0} V_{0} \log_{e} 2^{5} = 160 P_{0} V_{0} \log_{e} 2$

$(1) is the correct option$

Solution

1 P-4, Q-3, R-2, S-1

2 P-4, Q-3, R-1, S-2

3 P-3, Q-1, R-2, S-4

4 P-1, Q-3, R-2, S-4

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