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Ideal Gas Equation

One mole of a...

Question

One mole of an ideal gas undergoes a process $P = \frac{P_{0}}{1 + {(\frac{V_{0}}{V})}^{2}}$ Here $P_{0}$ and $V_{0}$ are constants. Change in temperature of the gas when volume is changed from $V = V_{0}$ to $V = 2 V_{0}$ is :-

- \frac{2 P_{0} V_{0}}{5 R}

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\frac{11 P_{0} V_{0}}{10 R}

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\frac{5 P_{0} V_{0}}{4 R}

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P_{0} V_{0}

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Solution

The correct option is B $\frac{11 P_{0} V_{0}}{10 R}$
Given
$P = \frac{P_{0}}{1 + (\frac{V_{0}}{V})^{2}}$
Volume changes V= $V_{0}$ and V=2 $V_{0}$
Solution
Case 1 V= $V_{0}$
$P = \frac{P_{0}}{1 + (\frac{V_{0}}{V})^{2}}$
Putting V= $V_{0}$
$P = P_{0} / 2$
$T = \frac{P V}{n R}$
$T = \frac{P V}{n R}$
Putting values
$T = \frac{P_{0} V_{0}}{2 n R}$

Case 2 V= $2 V_{0}$
$P = \frac{P_{0}}{1 + (\frac{V_{0}}{V})^{2}}$
Putting V= $2 V_{0}$
$P = 4 P_{0} / 5$
$T = \frac{P V}{n R}$
Putting values
$T = \frac{8 P_{0} V_{0}}{5 n R}$
Difference between temperatures
$T = \frac{11 P_{0} V_{0}}{10 n R}$

Suggest Corrections

Similar questions

P = \frac{P_{0}}{1 + (V_{0} / V)^{2}}

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

P = \frac{P_{0}}{1 + (V_{0} / V)^{2}}

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

P = \frac{P_{0}}{1 + (V_{0} / V)^{2}}

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

P = P_{0} {(1 + {(\frac{2 V_{0}}{V})}^{2}]}^{- 1},

P_{0}, V_{0}

V = V_{0}

V = 2 V_{0}

\frac{n P_{o} V_{0}}{R}

n

p = \frac{p_{0}}{1 + (V / V_{0})^{2}}

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