Question

Two pistons having low thermal conductivity divide an adiabatic container in three equal parts as shown. An ideal gas is present in the three parts A, B and C having initial pressures as shown and same temperatures. Now the pistons are released. Then the final equilibrium length of part A after long time will be

• A. $\frac{L}{8}$
• B. $\frac{L}{4}$
• C. $\frac{L}{6}$
• D. $\frac{L}{5}$

Answer 1

The correct option is B $\frac{L}{4}$

According to $PV=nRT$
Where $P$ = Pressure
$V$ = volume
$n$ = number of moles
$R$ = universal gas constant
$T$ = temperature
number of moles for A -
$n_A=\frac{PV}{RT}=\frac{PLA}{3RT}$
for $n_B-$
$n_B=\frac{2PLA}{3RT}$
for $n_c=\frac{PLA}{3RT}$
Then total number of moles-
$n_{total}=n_A+n_B+n_C$
$=\frac{PLA}{3RT}+\frac{2PLA}{3RT}+\frac{PLA}{3RT}$
$=\frac{PLA}{3RT}[1+2+1]=\frac{4PLA}{3RT}$
According to $n_{\text{total}}$ $n_A=\frac{1}{4}{n}_{\text{total}},n_B=\frac{1}{2}{n}_{\text{total}}$
$n_C=\frac{1}{4}{n}_{\text{total}}$
Since the temperature and pressure in all the parts will be equal at equilibrium,
$V\propto n$ or,
$L\propto n$
So, final equilibrium length of part A after long time will be -
since $n_A=\frac{1}{4}{n}_{\text{total}}$
Hence, $L^{\prime }=\frac{L}{4}$