MHT CET · Physics · Thermodynamics
Temperature remaining constant, the pressure of gas is decreased by \(20 \%\). The percentage change in volume
- A increases by \(29 \%\)
- B decreases by \(20 \%\)
- C increases by \(25 \%\)
- D decreases by \(25 \%\)
Answer & Solution
Correct Answer
(C) increases by \(25 \%\)
Step-by-step Solution
Detailed explanation
\(\text { Using, } \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \text { we get, } \)
\( \mathrm{PV}=\mathrm{P}^{\prime} \times \frac{80 \mathrm{~V}}{100} \Rightarrow \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{10}{8} \)
\( \therefore \frac{\mathrm{P}^{\prime}-\mathrm{P}}{\mathrm{P}} \times 100 =\left(\frac{10}{8}-1\right) \times 100 \)
\( =\left(\frac{2}{8} \times 100\right)=\frac{1}{4} \times 100=25 \%\)
\( \mathrm{PV}=\mathrm{P}^{\prime} \times \frac{80 \mathrm{~V}}{100} \Rightarrow \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{10}{8} \)
\( \therefore \frac{\mathrm{P}^{\prime}-\mathrm{P}}{\mathrm{P}} \times 100 =\left(\frac{10}{8}-1\right) \times 100 \)
\( =\left(\frac{2}{8} \times 100\right)=\frac{1}{4} \times 100=25 \%\)
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