A thermodynamic system goes from states (i) $P_{1}, V$ to $2 P_{1}, V$ (ii) $P_{1}, V_{1}$ to $P_{1}, 2 V_{1}$ . Then work done in the two cases is respectively

0, 0

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0, P_{1} V_{1}

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P_{1} V_{1}, 0

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P_{1} V_{1}, P_{1} V_{1}

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Solution

The correct option is A $0, P_{1} V_{1}$
(i) Case $\to v o l u m e = c o n s t a n t \Rightarrow$ $\int P d V = 0$
(ii) Case $\to P = c o n s t a n t$
$\Rightarrow$ $\int_{V_{1}}^{2 V_{1}} P d V = P \int_{V_{1}}^{2 V_{1}} d V = P V_{1}$

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	Column I		Column II
(A)	$P = p_{1} + p_{2} + p_{3} + . . . .$	(i)	Boyle's law
(B)	$P_{1} V_{1} = P_{2} V_{2} = P_{3} V_{3} = . . . .$	(ii)	Ideal gas equation
(C)	$(V - b) (P + \frac{a}{V^{2}}) = R T$	(iii)	Dalton's law of partial pressure
(D)	$P V = n R T$	(iv)	Equation for real gases

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