MHT CET · Physics · Kinetic Theory of Gases
A gas at pressure \(\mathrm{P}_0\) is contained in a vessel. If the masses of all the molecules are halved and their velocities are doubled, the resulting pressure would be equal to
- A \(4 \mathrm{P}_0\)
- B \(2 \mathrm{P}_0\)
- C \(\mathrm{P}_0\)
- D \(\frac{P_0}{2}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{P}_0\)
Step-by-step Solution
Detailed explanation
The pressure is given as \(\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{v}_{\mathrm{rms}}^2\)
\(\Rightarrow \mathrm{P} \propto \mathrm{mv}_{\mathrm{ms}}^2\)
\(\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{~m}_1 \mathrm{v}_1^2}\)
\(\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\frac{\mathrm{m}_1}{2}\left(2 \mathrm{v}_1\right)^2}{\mathrm{~m}_1 \mathrm{v}_1^2}=2\)
\(\therefore \quad \mathrm{P}_2=2 \mathrm{P}_0\) \(\ldots\left(\because \mathrm{P}_1=\mathrm{P}_0\right)\)
\(\Rightarrow \mathrm{P} \propto \mathrm{mv}_{\mathrm{ms}}^2\)
\(\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{~m}_1 \mathrm{v}_1^2}\)
\(\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\frac{\mathrm{m}_1}{2}\left(2 \mathrm{v}_1\right)^2}{\mathrm{~m}_1 \mathrm{v}_1^2}=2\)
\(\therefore \quad \mathrm{P}_2=2 \mathrm{P}_0\) \(\ldots\left(\because \mathrm{P}_1=\mathrm{P}_0\right)\)
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