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  • Figure shows a cylindrical tube of volume divided in two parts by a frictionless separato

Figure shows a cylindrical tube of volume \mathrm{V}_{0} divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gasses are filled in the two parts. When the separator is kept in the middle, the pressures are \mathrm{p_{1}} and \mathrm{p_{2}} in the left part and the right part respectively. The separator is slowly slide and is released at a position where it can stay in equilibrium. Find the volume of the two parts.

 

Option: 1

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Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

As the separator is conducting, the temperatures in the two parts will be the same. Suppose the common temperature is \mathrm{T} when the separator is in the middle. Let \mathrm{n}_{1} and \mathrm{n}_{2}  be the number of moles of the gas in the left part and the right part respectively. Using ideal gas equation,

\mathrm{and \quad \mathrm{p}_{2} \frac{\mathrm{V}_{0}}{2}=\mathrm{n}_{2} \mathrm{RT}}
\mathrm{p}_{1} \frac{\mathrm{V}_{0}}{2}=\mathrm{n}_{1} \mathrm{RT}
Thus, \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\quad \ldots(i)

The separator will stay in equilibrium at a position where the pressures on the two sides are equal. Suppose the volume of the left part is \mathrm{V}_{1}and of the right part is \mathrm{V}_{2} in this situation. Let the common pressure be p'. Also, let the common temperature in this situation be \mathrm{'T'}.

Using ideal gas equation
\mathrm{p}^{\prime} \mathrm{V}_{1}=\mathrm{n}_{1} \mathrm{RT}^{\prime}$ and $\quad \mathrm{p}^{\prime} \mathrm{V}_{2}=\mathrm{n}_{2} \mathrm{RT}^{\prime}

or, \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{p}_{\mathrm{I}}}{\mathrm{p}_{2}}\left [ using(i) \right ]
Also, \quad \mathrm{V}_{1}+\mathrm{V}_{2}=\mathrm{V}_{0}

Thus, \quad \mathrm{V}_{1}=\frac{\mathrm{p}_{1} \mathrm{~V}_{0}}{\mathrm{p}_{1}+\mathrm{p}_{2}}$ and $\mathrm{V}_{2}=\frac{\mathrm{p}_{2} \mathrm{~V}_{0}}{\mathrm{p}_{1}+\mathrm{p}_{2}}

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