Question

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P =P_{o} \left[1- \frac{1}{2} \left(\frac{V_{0}}{V}\right)^{2}\right] $ .Here $P_o$ and $V_o$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_o$ to $2 V_o$ .

• A. $\frac{1}{2} \frac{P_{o}V_{o}}{R} $
• B. $\frac{3}{4} \frac{P_{o}V_{o}}{R} $
• C. $\frac{5}{4} \frac{P_{o}V_{o}}{R} $
• D. $\frac{1}{4} \frac{P_{o}V_{o}}{R} $

Answer 1

The Correct Option is C

$P = P_{0} \left[ 1- \frac{1}{2} \left(\frac{V_{0}}{V}\right)^{2}\right] $ Pressure at $ V_{0} =P_{0} \left(1- \frac{1}{2}\right) = \frac{P_{0}}{2} $ Pressure at $ 2V_{0} = P_{0} \left(1- \frac{1}{2} \times\frac{1}{4}\right) = \frac{7}{8} P_{0} $ Temperature at $ V_{0} = \frac{\frac{P_{0}}{2} V_{0}}{nR} = \frac{P_{0}V_{0}}{2nR} $ Temperature at $ 2V_{0} = \frac{\left(\frac{7}{8} P_{0}\right)\left(2V_{0}\right)}{nR} = \frac{\frac{7}{4}P_{0}V_{0}}{nR } $ $ =\left(\frac{7}{4} - \frac{1}{2}\right) \frac{P_{0}V_{0}}{nR} $ Change in temperature$ = \frac{5}{4} \frac{P_{0}V_{0}}{nR} = \frac{5P_{0}V_{0}}{4R} $