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A gas expands obeying the relation as shown in the \mathrm{P} / \mathrm{V} diagram. The maximum temperature in this process is equal to

Option: 1

\mathrm{\frac{P_{0} V_{0}}{R}}


Option: 2

\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}


Option: 3

\mathrm{\frac{9 P_{0} V_{0}}{8 R}}


Option: 4

None of these


Answers (1)

best_answer

The equation of  \mathrm{PQ} is given as

\mathrm{\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{x}-\mathrm{x}_{1}}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}}

\mathrm{\Rightarrow \frac{\mathrm{P}-\mathrm{P}_{0}}{\mathrm{~V}-\mathrm{V}_{0}}=\frac{\mathrm{P}_{0} / 2-\mathrm{P}_{0}}{2 \mathrm{~V}_{0}-\mathrm{V}_{0}} }
\mathrm{\Rightarrow \frac{\mathrm{P}-\mathrm{P}_{0}}{\mathrm{~V}-\mathrm{V}_{0}}=-\frac{\mathrm{P}_{0}}{2 \mathrm{~V}_{0}} }
\mathrm{\Rightarrow 2 \mathrm{PV} \mathrm{V}_{0}-2 \mathrm{P}_{0} \mathrm{~V}=-\mathrm{P}_{0} \mathrm{~V}+\mathrm{P}_{0} \mathrm{~V}_{0} }
\mathrm{ \Rightarrow 2 \mathrm{P} \mathrm{V}_{0}+\mathrm{P}_{0} \mathrm{~V}=3 \mathrm{P}_{0} \mathrm{~V} }
\mathrm{ \Rightarrow \mathrm{P}=\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{2 \mathrm{~V}_{0}}=\frac{3}{2} \mathrm{P}_{0}-\frac{\mathrm{P}_{0}}{2 \mathrm{~V}_{0}} \mathrm{~V} }

\mathrm{ \because \mathrm{PV}=\mathrm{RT} }
\mathrm{ \frac{\mathrm{RT}}{\mathrm{V}}=\frac{3}{2} \mathrm{P}_{0}-\frac{\mathrm{P}_{0} \mathrm{~V}}{2 \mathrm{~V}_{0}}}
\mathrm{ \mathrm{T}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}}{\mathrm{R}}-\frac{\mathrm{P}_{0} \mathrm{~V}^{2}}{2 \mathrm{~V}_{0} \mathrm{R}}}\quad \ldots (1)

\mathrm{\frac{d T}{d V}=0\, \&\, \frac{d^{2} T}{d V^{2}}<0 for T_{\max }}

\mathrm{\Rightarrow \frac{\mathrm{dT}}{\mathrm{dV}}=\frac{3 \mathrm{P}_{0}}{2 \mathrm{R}}-\frac{\mathrm{P}_{0} \mathrm{~V}}{\mathrm{~V}_{0} \mathrm{R}}=0}
\mathrm{\Rightarrow \mathrm{V}=\frac{3 \mathrm{~V}_{0}}{2}}

substituting this value of \mathrm{V}in equation (1)
\mathrm{T}_{\max }=\frac{9 \mathrm{P}_{0} \mathrm{~V} 0}{8 \mathrm{R}}
\therefore(\mathrm{C})

Posted by

Suraj Bhandari

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