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Two gases occupy two containers $A$ and $B$ the gas in $\mathrm{A},$ of volume $0.10 \mathrm{~m}^{3},$ exerts a pressure of $1.40 \mathrm{MPa}$ and that in $\mathrm{B}$ of volume $0.15 \mathrm{~m}^{3}$ exerts a pressure $0.7 \mathrm{MPa}$. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)
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The correct answer is:
0.98
We know that $\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}=\mathrm{n}_{\mathrm{A}} \mathrm{RT}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{n}_{\mathrm{B}} \mathrm{RT}$
$\begin{array}{ll} & \text { and } P_{f}\left(V_{A}+V_{B}\right)=\left(n_{A}+n_{B}\right) R T \\ & P_{f}\left(V_{A}+V_{B}\right)=P_{A} V_{A}+P_{B} V_{B} \\ \therefore & P_{f}=\left(\frac{P_{A} V_{A}+P_{B} V_{B}}{V_{A}+V_{B}}\right) \\ & =\frac{1.4 \times 0.1+0.7 \times 0.15}{0.1+0.15} \mathrm{MPa}=0.98 \mathrm{MPa}\end{array}$
$\begin{array}{ll} & \text { and } P_{f}\left(V_{A}+V_{B}\right)=\left(n_{A}+n_{B}\right) R T \\ & P_{f}\left(V_{A}+V_{B}\right)=P_{A} V_{A}+P_{B} V_{B} \\ \therefore & P_{f}=\left(\frac{P_{A} V_{A}+P_{B} V_{B}}{V_{A}+V_{B}}\right) \\ & =\frac{1.4 \times 0.1+0.7 \times 0.15}{0.1+0.15} \mathrm{MPa}=0.98 \mathrm{MPa}\end{array}$
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