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  • Two identical containers A and B fitted with frictionless pistons contain the same idal gas at the same temperature and the same volume V. The mass of the gas in A is <img alt="\mathrm{m_A}" src="h

Two identical containers A and B fitted with frictionless pistons contain the same idal gas at the same temperature and the same volume V. The mass of the gas in A is \mathrm{m_A} and that in B is \mathrm{m_B}. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in pressure in A and B are found to be P and 1.5 P respectively. Then

Option: 1

\mathrm{4 m_A=9 m_B}


Option: 2

\mathrm{2 m_A=3 m_B}


Option: 3

\mathrm{3 m_A=2 m_B}


Option: 4

\mathrm{9 m_A=4 m_B}


Answers (1)

best_answer

The equation of state for an ideal gas of mass m and molecular mass M is

\mathrm{P V=\frac{m}{M} R T}           .... [i]

For an isothermal process, T = constant. Differentiating (i) partially at constant T, we get

\mathrm{\begin{aligned} & P \Delta V+V \Delta P=0 \\ & \text { Or } \Delta P=-P \frac{\Delta V}{V} \end{aligned}}                   ....[ii]

\mathrm{\text { From (i), } P=\frac{m R T}{M V} \text {. Using this in (ii), we get }}

\mathrm{\begin{aligned} & \Delta P=-\frac{m R T}{M V}(\because \Delta V=2 V-V=V) \\ & \therefore \Delta P_A=-\frac{m_A R T}{M V} \text { and } \Delta P_B=-\frac{m_R R T}{M V} \end{aligned}}

\mathrm{\text { Hence } \frac{\Delta P_A}{\Delta P_B}=\frac{m_A}{m_B}}

\mathrm{\text { Given } \Delta P_B=1.5 \Delta P_A \text {. Therefore, } \frac{1}{1.5}=\frac{m_A}{m_B}}

\mathrm{\text { Or } 3 m_A=2 m_B}

 

 

 

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