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(D) For an isothermal process,the temperature remains constant,so Boyle's Law applies: $P_1 V_1 = P_2 V_2$. Given that the pressure is decreased by $1

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increase by $11 \%$

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(D) For an isothermal process,the temperature remains constant,so Boyle's Law applies: $P_1 V_1 = P_2 V_2$. Given that the pressure is decreased by $10 \%$,the new pressure $P_2$ is: $P_2 = P_1 - 0.10 P_1 = 0.9 P_1 = \frac{9}{10} P_1$. Substituting this into the equation: $P_1 V_1 = (\frac{9}{10} P_1) V_2$. Solving for $V_2$: $V_2 = \frac{10}{9} V_1 \approx 1.111 V_1$. The change in volume is $\Delta V = V_2 - V_1 = 1.111 V_1 - V_1 = 0.111 V_1$. Expressed as a percentage,the volume increases by approximately $11.1 \%$,which is closest to $11 \%$.

If the pressure of an ideal gas is decreased by $10 \%$ isothermally,then its volume will

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