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Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume \mathrm{V}. The mass of gas \mathrm{A} is \mathrm{m_{A}} and that in \mathrm{B} is \mathrm{\mathrm{m}_{\mathrm{B}}}.The gas in each cylinder is now allowed to expand isothermally to the same final volume 2 \mathrm{~V}.The change in the pressure in \mathrm{A} and \mathrm{B} are found to be \Delta \mathrm{P respectively. Then

Option: 1

4 \mathrm{~m}_{\mathrm{A}}=9 \mathrm{~m}_{\mathrm{B}}


Option: 2

\mathrm{2 m_{A}=3 m_{B}}


Option: 3

3 \mathrm{~m}_{\mathrm{A}}=2 \mathrm{~m}_{\mathrm{B}}


Option: 4

9 \mathrm{~m}_{\mathrm{A}}=4 \mathrm{~m}_{\mathrm{B}}


Answers (1)

best_answer

For gas in \mathrm{A}, \mathrm{P}_{1}=\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{V}_{1}}

\mathrm{ \mathrm{P}_{2}=\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{V}_{2}} }
\mathrm{ \therefore \quad \Delta \mathrm{P}=\mathrm{P}_{1}-\mathrm{P}_{2}=\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \mathrm{m}_{\mathrm{A}}\left(\frac{1}{\mathrm{~V}_{1}}-\frac{1}{\mathrm{~V}_{2}}\right) }

Putting \mathrm{\quad \mathrm{V}_{1}=\mathrm{V} \: and \: \mathrm{V}_{2}=2 \mathrm{~V} }
We get \mathrm{\quad \Delta \mathrm{P}=\frac{\mathrm{RT}}{\mathrm{M}} \frac{\mathrm{m}_{\mathrm{A}}}{2 \mathrm{~V}}}
Similarly for Gas in\mathrm{B, 1.5 \Delta P=\left(\frac{R T}{M}\right) \frac{m_{B}}{2 V}}

From eq. (I) and (II) we get\mathrm{2 \mathrm{~m}_{\mathrm{B}}=3 \mathrm{~m}_{\mathrm{A}}}

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manish

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