Question

A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has a pressure P and volume 5 V and the other part has a pressure 8 P and volume V. The piston is now left free. Find the new pressures and volumes for the adiabatic and isothermal processes. For this gas $\gamma =\mathrm{1.5.}$

Answer 1

Final pressure will be same on both sides. Let it be P, with volume V, on the left side and (6V_0-v) on the right side.

Case 1: Adiabatic Process

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

For the gas enclosed in the left chamber,

$P_0\times (5V_0{)}^{\gamma }=P(V{)}^{\gamma }.........\left(i\right)$

while for the gas in the right cylinder,

$8P_0(V_0{)}^{\gamma }=P(6V_0-V{)}^{\gamma }.........\left(ii\right)$

Dividing (i) and (ii), we get

${\left(\frac{6V_0-V}{V}\right)}^{\gamma }=\frac{8}{5^{\gamma }}$

or, $\frac{6V_0}{V}=1+\frac{4}{5}$

i.e., $V=\frac{10}{3}{V}_0$

substituting it in equation(ii),

$P=P_0{\left(\frac{5V\times 3}{10V}\right)}^{\frac{3}{2}}=\frac{3\sqrt{3}}{2\sqrt{2}}{P}_0=1.84P_0$

$P=1.84P{)}_0,V=\frac{10}{3}{V}_0,(6V_0-V)=\frac{8}{3}{V}_0$

Case 2: Isothermal process

For the gas enclosed in the left chamber,

$P_0\times 5V_0=PV$.......(i)

while for the gas in the right cylinder

$8P_0\times V_0=P(6V_0-V)$........(ii)

After solving, we get

$V=\frac{30}{13}{V}_0$ and $P=\frac{13}{6}{P}_0$

and $(6V_0-V)=\frac{48}{13}{V}_0$