One mole of an ideal gas undergoes a process.P=P01+V0V2Here P0 and V0 are constants. Change in temperature of the gas when volume is changed from V=V0 to V=2V0is

At V=V0:P=P02Ti=PVnR=P02V0R=P0V02R(n=1) and at V=2V0:P=4P05∴ Tf=PVnR=2V04P05R=8P0V05R∴ ΔT=Tf−Ti=85−12P0V0R

One mole of an ideal gas undergoes a process.P=P01+V0V2Here P0 and V0 are constants. Change in temperature of the gas when volume is changed from V=V0 to V=2V0is

At V=V0:P=P02Ti=PVnR=P02V0R=P0V02R(n=1) and at V=2V0:P=4P05∴ Tf=PVnR=2V04P05R=8P0V05R∴ ΔT=Tf−Ti=85−12P0V0R

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One mole of an ideal gas undergoes a process. $P = \frac{P_{0}}{1 + {(\frac{V_{0}}{V})}^{2}}$ Here P₀ and V₀ are constants. Change in temperature of the gas when volume is changed from $V = V_{0} to V = 2 V_{0}$ is

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$P_{0} V_{0}$

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

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

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

answer is B.

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Detailed Solution

At $V = V_{0} : P = \frac{P_{0}}{2}$
$\begin{array}{l} T_{i} = \frac{PV}{nR} = \frac{(\frac{P_{0}}{2}) V_{0}}{R} = \frac{P_{0} V_{0}}{2 R} (n = 1) \\ and at V = 2 V_{0} : P = \frac{4 P_{0}}{5} \\ ∴ T_{f} = \frac{PV}{nR} = \frac{(2 V_{0}) (\frac{4 P_{0}}{5})}{R} = \frac{8 P_{0} V_{0}}{5 R} \\ ∴ ΔT = T_{f} - T_{i} = (\frac{8}{5} - \frac{1}{2}) \frac{P_{0} V_{0}}{R} \end{array}$