MHT CET · Physics · Thermodynamics
If the pressure of an ideal gas is decreased by \(10 \%\) isothermally, then its volume will
- A decrease by \(8 \%\)
- B decrease by \(9 \%\)
- C increase by \(8 \%\)
- D increase by \(11 \%\)
Answer & Solution
Correct Answer
(D) increase by \(11 \%\)
Step-by-step Solution
Detailed explanation
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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