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One mole of a...

Question

One mole of an ideal gas is heated slowly so that it goes from the $P - V$ state $(P_{i}, V_{i})$ to $(3 P_{i}, 3 V_{i})$ in such a way that the pressure of the gas is directly proportional to the volume. How much work is done on the gas in the process?

A

4 P_{i} V_{i}

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B

16 P_{i} V_{i}

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C

- 9 P_{i} V_{i}

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D

- 4 P_{i} V_{i}

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Solution

The correct option is D $- 4 P_{i} V_{i}$
Given $P \propto V$
i.e $P = k V$
So, during the heating process, $P = (\frac{P_{i}}{V_{i}}) V$
[since $P_{i} = k V_{i}$ ]
Work done on the gas is
$W = - \int_{V_{i}}^{V_{f}} P d V = - \int_{V_{i}}^{3 V_{i}} (\frac{P_{i}}{V_{i}}) V d V$
$W = - (\frac{P_{i}}{V_{i}}) {[\frac{V^{2}}{2}]}_{V_{i}}^{3 V_{i}} = - \frac{P_{i}}{2 V_{i}} (9 V_{i}^{2} - V_{i}^{2})$
$W = - 4 P_{i} V_{i}$

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