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Thermodynamic Processes

A gas is foun...

Question

A gas is found to obey the law $p^{2} V =$ constant . The initial temperature and volume are $T_{0}$ and $V_{0}$ . If the gas expands to a volume $3 V_{0}$ , its final temperature becomes

\frac{T_{0}}{3}

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\frac{T_{0}}{\sqrt{3}}

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\sqrt{3}

T_{0}

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None of these

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Solution

The correct option is C $\sqrt{3}$ $T_{0}$
Since $p^{2} V = c o n s t a n t$
$(\frac{R T}{V})^{2} V = c o n s t a n t$
$\frac{T^{2}}{V} = c o n s t a n t$
$\frac{T_{2}}{T_{1}} = (\frac{V_{2}}{V_{1}})^{1 / 2}$
$T_{2} = T_{0}$ $\sqrt{3}$

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A gas is found to obey the law p2V= constant . The initial temperature and volume are T0 and V0. If the gas expands to a volume 3V0 , its final temperature becomes

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A gas is found to obey the law $p^{2} V =$ constant . The initial temperature and volume are $T_{0}$ and $V_{0}$ . If the gas expands to a volume $3 V_{0}$ , its final temperature becomes

The correct option is C $\sqrt{3}$ $T_{0}$
Since $p^{2} V = c o n s t a n t$
$(\frac{R T}{V})^{2} V = c o n s t a n t$
$\frac{T^{2}}{V} = c o n s t a n t$
$\frac{T_{2}}{T_{1}} = (\frac{V_{2}}{V_{1}})^{1 / 2}$
$T_{2} = T_{0}$ $\sqrt{3}$