At a constant temperature,by what percentage | vedclass

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(D) According to Boyle's Law,for a given mass of gas at constant temperature,$P_1V_1 = P_2V_2$.Let the initial pressure be $P_1 = P$ and initial volum

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$11$

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(D) According to Boyle's Law,for a given mass of gas at constant temperature,$P_1V_1 = P_2V_2$.Let the initial pressure be $P_1 = P$ and initial volume be $V_1 = V$.The final volume is $V_2 = V - 0.1V = 0.9V$.Using $P_1V_1 = P_2V_2$,we get $P 	imes V = P_2 	imes 0.9V$.Therefore,$P_2 = \frac{P}{0.9} = \frac{10}{9}P$.The change in pressure is $\Delta P = P_2 - P = \frac{10}{9}P - P = \frac{1}{9}P$.The percentage increase in pressure is $\frac{\Delta P}{P} 	imes 100 = \frac{1/9P}{P} 	imes 100 = \frac{100}{9} \approx 11.11\%$.

At a constant temperature,by what percentage $(\%)$ must the pressure of a given mass of gas be increased so that its volume decreases by $10\%$ (in $.1$)?

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