Question

A vertical cylinder containing some ideal gas is closed
at both ends. A frictionless piston is fitted in the cylinder
such that it divides the cylinder into two equal parts at
300K. Moles of gas in one part is double than that in
other part. To what temperature (in Kelvin) the cylinder
should be heated or cooled such that the volume of
upper part becomes three times that of lower part?

Answer 1

Given:

Cyclinder is divided into two parts at 300$k$

One mole of gas present in each part

Volume of upper part is double than that lower part

To Find:

At what temperature Volume of Upper part becomes 3 times of lower part

Solution:

$P_{U}$ = Pressure of upper part

$P_{L}$ = Pressure of lower part

$P_{P}$ = Pressure of piston

At Equillibrium

$P_{L} = P_{U} + P_{P}$

$\dfrac{nRT}{V_{1} } = \dfrac{nRT}{V_{2} } + P_{P}$

$\dfrac{3R 300}{1V } = \dfrac{3R300}{2V } + P_{P}$

Below is first Equation

$P_{P} = \dfrac{900R}{2V}$

$\dfrac{nRT}{V_{1} } = \dfrac{nRT}{V_{2} } + P_{P}$

$\dfrac{4R T}{1V } = \dfrac{4RT}{3V } + P_{P}$

Below is Second Equation

$P_{P} = \dfrac{8RT}{3V}$

By Equating these two equation

Temperature is 168.75$K$