In order to increase the volume of a gas by 1 | vedclass

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(D) According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$. Let the initial volume be $V_1 = V$. The new volume $V_2 = V + 0.10V = 1.10V$.

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decreased by $10 \%$

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(D) According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$. Let the initial volume be $V_1 = V$. The new volume $V_2 = V + 0.10V = 1.10V$. Substituting into the equation: $P_1V = P_2(1.10V)$. $P_2 = P_1 / 1.10 \approx 0.9091 P_1$. The change in pressure is $P_2 - P_1 = 0.9091 P_1 - P_1 = -0.0909 P_1$. This represents a decrease of approximately $9.09 \%$. Given the options provided,the closest value is a decrease of $10 \%$,which corresponds to option $(D)$.

In order to increase the volume of a gas by $10 \%$ at constant temperature,the pressure of the gas should be:

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