Question

An adiabatic cylinder of volume $V_0$ is divided by an adiabatic piston in two equal compartments. Both the compartments are filled with ideal monoatomic gas at pressure and temperature $(P_0,T_0)\text{and}(32P_0,32T_0)$ respectively. With the help of external agent, piston is moved slowly such that it is in equilibrium, then

• A. Final pressure of gas is $P_0{\left(\frac{9}{2}\right)}^{3/5}$
• B. Final pressure of gas is $P_0{\left(\frac{9}{2}\right)}^{5/3}$
• C. Ratio of temperature in right compartment to left compartment is 8
• D. Ratio of temperature in right compartment to left compartment is 4

Answer 1

Answer: (B), (C)

Ratio of temperature in right compartment to left compartment is 8


$\therefore =\frac{5}{3}$
For left part : $P_0{\left(\frac{V_0}{2}\right)}^{\gamma }=P{\left(V_1\right)}^{\gamma }$
For right part : $32P_0{\left(\frac{V_0}{2}\right)}^{\gamma }=P{\left(V_2\right)}^{\gamma }$
$\therefore 32={\left(\frac{V_2}{V_1}\right)}^{\gamma }\Rightarrow 32={\left(\frac{V_2}{V_1}\right)}^{5/3}$
$\therefore \left(\frac{V_2}{V_1}\right)={\left(32\right)}^{3/5}$
$\therefore V_2=8V_1\text{Also}{V}_1+V_2=V_0$
$\therefore V_1+8V_1=V_0\Rightarrow V_1=\frac{V_0}{9}\text{~and}{V}_2=\frac{8V_0}{9}$
$\therefore$Final pressure $P=\frac{P_0{\left(V_0\right)}^{5/3}{\left(9\right)}^{5/3}}{{\left(2\right)}^{5/3}{\left(V_0\right)}^{5/3}}$
$\therefore P=P_0{\left(\frac{9}{2}\right)}^{5/3}$
Now $T_0{\left(\frac{V_0}{2}\right)}^{\gamma -1}=T_1{\left(V_1\right)}^{\gamma -1}\Rightarrow T_1=T_0{\left(\frac{81}{4}\right)}^{1/3}$
And $32T_0{\left(\frac{V_0}{2}\right)}^{\gamma -1}=T_2{\left(\frac{8V_0}{9}\right)}^{\gamma -1}$
$\Rightarrow T_2=8T_0{\left(\frac{81}{4}\right)}^{1/3}$
$\therefore \frac{T_2}{T_1}=8$