InterviewSolution
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InterviewSolution
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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 velocity V. The mass of the gas in A is `m_A,` and that in B is `m_B`. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. ThenA. `4m_(A)=9m_(B)`B. `2m_(A)=3m_(B)`C. `3m_(A)=2m_(B)`D. `9m_(A)=4m_(B)` |
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Answer» Correct Answer - C Process is isothermal. Therefore, T = constant. Volume is increasing, therefore, pressure will decrease `(Pprop(1)/(V)).` In chamber `A,Deltap=(pA)i-(pA)f=(n_(A)RT)/(V)-(n_(A)RT)/(2V).` `=(n_(A)RT)/(2V)" "...(i)` In chamber `B,1.5Deltap=(pa),-(p_(B))f=(n_(B)RT)/(V)-(n_(B)RT)/(2V)` `=(n_(B)RT)/(2V)" "...(ii)` From eqs. (i) and (ii), `(n_(A))/(n_(B))=(1)/(1.5)=2/3,(m_(A)//M)/(m_(B)//M)=2/3or (m_(A))/(m_(B))=2/3therefore3m_(A)=2m_(B)` |
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