Two ideal gases $A$ and $B$ having the same temperature $\mathrm{T}$, same pressure $\mathrm{P}$ and same…

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
  1. $\frac{P}{2}$
  2. $\mathrm{P}$
  3. $4 \mathrm{P}$
  4. $2 \mathrm{P}$

Solution

Initially $\mathrm{P}_{1}=\mathrm{P}$ $\mathrm{V}_{1}=\mathrm{V}+\mathrm{V}=2 \mathrm{~V}$ Finally $P_{2}=P_{1} V_{2}=\frac{V}{2}$ $\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$ $\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{P} \times 2 \mathrm{~V}}{\frac{\mathrm{V}}{2}}=4 \mathrm{P}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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