Solveeit Logo

Question

Chemistry Question on Ideal gas equation

One mole of an ideal gas passes through a process where pressure and volume obey the relation P=Po[112(V0V)2]P =P_{o} \left[1- \frac{1}{2} \left(\frac{V_{0}}{V}\right)^{2}\right] .Here PoP_o and VoV_o are constants. Calculate the change in the temperature of the gas if its volume changes from VoV_o to 2Vo2 V_o .

A

12PoVoR\frac{1}{2} \frac{P_{o}V_{o}}{R}

B

34PoVoR\frac{3}{4} \frac{P_{o}V_{o}}{R}

C

54PoVoR\frac{5}{4} \frac{P_{o}V_{o}}{R}

D

14PoVoR\frac{1}{4} \frac{P_{o}V_{o}}{R}

Answer

54PoVoR\frac{5}{4} \frac{P_{o}V_{o}}{R}

Explanation

Solution

P=P0[112(V0V)2]P = P_{0} \left[ 1- \frac{1}{2} \left(\frac{V_{0}}{V}\right)^{2}\right] Pressure at V0=P0(112)=P02V_{0} =P_{0} \left(1- \frac{1}{2}\right) = \frac{P_{0}}{2} Pressure at 2V0=P0(112×14)=78P02V_{0} = P_{0} \left(1- \frac{1}{2} \times\frac{1}{4}\right) = \frac{7}{8} P_{0} Temperature at V0=P02V0nR=P0V02nRV_{0} = \frac{\frac{P_{0}}{2} V_{0}}{nR} = \frac{P_{0}V_{0}}{2nR} Temperature at 2V0=(78P0)(2V0)nR=74P0V0nR2V_{0} = \frac{\left(\frac{7}{8} P_{0}\right)\left(2V_{0}\right)}{nR} = \frac{\frac{7}{4}P_{0}V_{0}}{nR } =(7412)P0V0nR=\left(\frac{7}{4} - \frac{1}{2}\right) \frac{P_{0}V_{0}}{nR} Change in temperature=54P0V0nR=5P0V04R= \frac{5}{4} \frac{P_{0}V_{0}}{nR} = \frac{5P_{0}V_{0}}{4R}