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

Option: 1

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


 


Option: 2

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


Option: 3

In the T-V graph, the path \mathrm{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} / \mathrm{n} 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 \mathrm{P}^{\prime}=\mathrm{P}+\frac{\mathrm{P}}{2}\left(\frac{\mathrm{V}^{\prime}}{-\mathrm{V}}+1\right) }

\mathrm{\therefore \mathrm{P}^{\prime}=-\frac{\mathrm{P}}{2 \mathrm{~V}} \mathrm{~V}^{\prime}+\frac{3}{2} \mathrm{P} }                ...........(1)

\mathrm{\therefore \frac{\mathrm{nRT}}{\mathrm{V}^{\prime}}=-\frac{\mathrm{P}}{2 \mathrm{~V}} \mathrm{~V}^{\prime}+\frac{3}{2} \mathrm{P} }

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

\mathrm{\therefore } Graph of \mathrm{\mathrm{T} \: vs \: \mathrm{V}^{\prime} } is a parabola for the process \mathrm{\mathrm{AB} }

\mathrm{ \frac{\mathrm{dT}}{\mathrm{dV}}=-\left(\frac{\mathrm{P}}{2 \mathrm{nRV}}\right) 2 \mathrm{~V}^{\prime}+\frac{3 \mathrm{P}}{2 \mathrm{nR}}=0 }

\mathrm{ \Rightarrow \quad \frac{2 \mathrm{~V}^{\prime}}{\mathrm{V}}=3 ; \mathrm{V}^{\prime}=\left(\frac{3}{2} \mathrm{~V}\right) }

\mathrm{ \therefore \quad \frac{\mathrm{d}^2 \mathrm{~T}}{\mathrm{dV}^{\prime 2}}=-\frac{\mathrm{P}}{\mathrm{nRV}} }

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







 




 

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Gunjita

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