If pressure of gas is increased by 5% then find the % decrease in volume at constant temperature

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7.44 %

8.59 %

4.77 %

5.89 %

answer is A.

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Detailed Solution

$\begin{array}{l} P_{2} = P_{1} + \frac{5}{100} P_{1} = \frac{105 P_{1}}{100} \end{array}$

$P_{1} V_{1} = P_{2} V_{2}$

$P_{1} V_{1} = \frac{105 P_{1} V_{2}}{100} \Rightarrow V_{2} = \frac{100 V_{1}}{105}$

$% change in volume = \frac{Δ V}{V_{1}} \times 100 = \frac{V_{2} - V_{1}}{V_{1}} \times 100$

$\Rightarrow (\frac{\frac{100 V_{1}}{100} - V_{1}}{V_{1}}) \times 100 = - \frac{500}{105} % = - 4.77 % [-Ve sign indicates decrease in volume]$

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