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  • An ideal gas is taken from the state A (pressure  and volume V) to the state B (pressure P/2 and

An ideal gas is taken from the state A (pressure \mathrm{p}  and volume V) to the state B (pressure P/2 and volume 2 V) along the straight line path in the \mathrm{P-V} graph. Select correct statement from the following :

Option: 1

The workdone by the gas in the process A to B exceeds the work that would be done by it, if the system was taken from A to B along an isothermal


Option: 2

In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases


Option: 3

In the T-V graph, the path AB becomes a part of a parabola


Option: 4

All of the above


Answers (1)

best_answer

Work = Area under curve

        \mathrm{=\frac{1}{2}\left(\frac{3}{2} \mathrm{P}\right) \mathrm{V}=\frac{3}{4} \mathrm{PV}}
       \mathrm{=\frac{3}{2} \mathrm{nRT}}

\mathrm{\mathrm{W}_{\text {Isothermal }} \mathrm{nRT} \ln 2}


\therefore  (a) is correct.

\mathrm{\text{Equation of line}\, \mathrm{AB}\, is \, \frac{\mathrm{P}-\mathrm{P}^{\prime}}{\mathrm{P}-\mathrm{P} / 2}=\frac{\mathrm{V}-\mathrm{V}^{\prime}}{\mathrm{V}-2 \mathrm{~V}}}
\mathrm{\Rightarrow \frac{\mathrm{P}-\mathrm{P}^{\prime}}{\mathrm{P} / 2}=\frac{\mathrm{V}-\mathrm{V}^{\prime}}{-\mathrm{V}}}
\mathrm{\therefore \quad \mathrm{P}^{\prime}=\mathrm{P}+\frac{\mathrm{P}}{2}\left(\frac{\mathrm{V}^{\prime}}{-\mathrm{V}}+1\right)}
\mathrm{\therefore \quad \mathrm{P}^{\prime}=-\frac{\mathrm{P}}{2 \mathrm{~V}} \mathrm{~V}^{\prime}+\frac{3}{2} \mathrm{P}\quad \ldots(1)}

\therefore \quad \frac{\mathrm{nRT}}{\mathrm{V}^{\prime}}=-\frac{\mathrm{P}}{2 \mathrm{~V}} \mathrm{~V}^{\prime}+\frac{3}{2} \mathrm{P}
\therefore \quad \mathrm{T}=\frac{-\mathrm{P}}{2 \mathrm{nRV}} \mathrm{V}^{\prime 2}+\frac{3}{2} \frac{\mathrm{P}}{\mathrm{nR}} \mathrm{V}^{\prime}\quad \ldots(2)

\therefore$ Graph of $\mathrm{T}$ vs $\mathrm{V}^{\prime}$ is a parabola for the process $\mathrm{AB}
\frac{\mathrm{dT}}{\mathrm{dV}^{\prime}}=-\left(\frac{\mathrm{P}}{2 \mathrm{nRV}}\right) 2 \mathrm{~V}^{\prime}+\frac{3 \mathrm{P}}{2 \mathrm{nR}}=0
\Rightarrow \quad \frac{2 \mathrm{~V}^{\prime}}{\mathrm{V}}=3 ; \mathrm{V}^{\prime}=\left(\frac{3}{2} \mathrm{~V}\right)
\therefore \quad \frac{\mathrm{d}^{2} \mathrm{~T}}{\mathrm{dV}^{\prime 2}}=-\frac{\mathrm{P}}{\mathrm{nRV}}

\Rightarrow \quad \mathrm{T}_{\max } \text { at } 1.5 \mathrm{~V} \text { then start decreasing. }


 

Posted by

shivangi.bhatnagar

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