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If the pressure of an ideal gas is decreased by $10 \%$ isothermally, then its volume will
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Verified Answer
The correct answer is:
increase by $11 \%$
For isothermal process, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$$
\begin{aligned}
& \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{P}_1}{10}=\frac{9}{10} \mathrm{P}_1 \\
& \therefore \mathrm{P}_1 \mathrm{~V}_1=\frac{9}{10} \mathrm{P}_1 \mathrm{~V}_2 \\
& \therefore \mathrm{V}_2=\frac{10}{9} \mathrm{~V}_1 \\
& =1.11 \mathrm{~V}_1 \\
& =\mathrm{V}_1+0.11 \mathrm{~V}_1
\end{aligned}
$$
Volume increase by $11 \%$
$$
\begin{aligned}
& \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{P}_1}{10}=\frac{9}{10} \mathrm{P}_1 \\
& \therefore \mathrm{P}_1 \mathrm{~V}_1=\frac{9}{10} \mathrm{P}_1 \mathrm{~V}_2 \\
& \therefore \mathrm{V}_2=\frac{10}{9} \mathrm{~V}_1 \\
& =1.11 \mathrm{~V}_1 \\
& =\mathrm{V}_1+0.11 \mathrm{~V}_1
\end{aligned}
$$
Volume increase by $11 \%$
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