MHT CET · Physics · Kinetic Theory of Gases
Two ideal gases \(A\) and \(B\) having the same temperature \(\mathrm{T}\), same pressure \(\mathrm{P}\) and
same volume \(\mathrm{V}\), are mixed together. If the temperature of mixture is kept constant
and the volume occupied by the mixture is reduced to \(\frac{\mathrm{V}}{2}\), then the pressure of the
mixture will become
- A \(\frac{P}{2}\)
- B \(\mathrm{P}\)
- C \(4 \mathrm{P}\)
- D \(2 \mathrm{P}\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{P}\)
Step-by-step Solution
Detailed explanation
Initially \(\mathrm{P}_{1}=\mathrm{P}\)
\(\mathrm{V}_{1}=\mathrm{V}+\mathrm{V}=2 \mathrm{~V}\)
Finally \(P_{2}=P_{1} V_{2}=\frac{V}{2}\)
\(\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}\)
\(\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{P} \times 2 \mathrm{~V}}{\frac{\mathrm{V}}{2}}=4 \mathrm{P}\)
\(\mathrm{V}_{1}=\mathrm{V}+\mathrm{V}=2 \mathrm{~V}\)
Finally \(P_{2}=P_{1} V_{2}=\frac{V}{2}\)
\(\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}\)
\(\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{P} \times 2 \mathrm{~V}}{\frac{\mathrm{V}}{2}}=4 \mathrm{P}\)
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