Question

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

• A. 2.84P
• B. 1.84P
• C. 1.48P
• D. 8.14P

Answer 1

The correct option is B 1.84P

Suppose, in isothermal process, the new pressure be P´ and the change in volume be v. Then, by Boyle's law,
we have $P\times 5V=P^{\prime }(5V-V)$ and
$8P\times V=P^{\prime }(V+v)$
Adding and subtracting these equations, we get :
$P^{\prime }=\frac{13}{6P}$ and $v=\left[\left(\frac{35}{13}\right)V\right]$
$\therefore$ New volumes are :
$V_1=5V-v=\left(\frac{30}{13}\right)V$ and
$V_2=V+v\left(\frac{48}{13}\right)V$
Let, in the adiabatic process, the new pressure be $P_1$ and change in volume be v. Then by poisson's law, we get
$P\left(5V\right)\gamma =P_1(5V-v)\gamma and8P\left(V\right)\gamma =P_1(V+v)\gamma \text{Dividing:}\frac{(5{)}^{\gamma }}{8}=\frac{5V-v}{V+v}or,\frac{5V-v}{V+v}=\frac{5}{8^{1/\gamma }}or,\frac{5V-v}{V+v}=\frac{5}{(8{)}^{2/3}}=\frac{5}{4}orv=\frac{5}{3}V\therefore \text{New volumes are:}$
$V_1=5V-v=\frac{10}{3}Vand{V}_2=V+v=\frac{8}{3}V$
Now, from the first equation of the adiabatic process, we get
$P_1=P\frac{5V}{\frac{10V}{3}}=P\sqrt{\frac{27}{8}}=1.84P$