Question

Figure shows an adiabatic cylindrical tube of volume $V_o$ divided in two parts by a frictionless adiabatic seperator. Initially, the seperator is kept in the middle, an ideal gas at pressure $P_1$ and temperature $T_1$ is injected into the left part and another ideal gas at pressure $P_2$ and temperature $T_2$ is injecteed into the right part. $C_p/C_v=\gamma$ is the same for both the gases. The seperator is slid slowely and is released at a position where it can stay in equlibrium. Find

(a) The volumes of the two parts.

(b) The heat given to the gas in the left part and

(c) The final common pressure of the gases.

Answer 1

For an adiabatic process, $P{V}^{\gamma }$ = Constant

So $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$ ....(i)

According to the problem

$V_1+V_2=V_o$

Then the equation (i)

$P_1{V}_2^{\gamma }=P_2(V_o-V_1{)}^{\gamma }$

${\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma }}=\frac{V_o-V_1}{V_1}$

or $V_1-P_1^{\frac{1}{\gamma }}=V_o-P_2^{\frac{1}{\gamma }}-V_1{P}_2^{\frac{1}{\gamma }}$

or $V_1(P_1^{\frac{1}{\gamma }}+P_2^{\frac{1}{\gamma }})=V_o{P}_2^{\frac{1}{\gamma }}$

or $V_1=\frac{P_2^{\frac{1}{\gamma }}{V}_o}{P_1^{\frac{1}{\gamma }}+P_2^{\frac{1}{\gamma }}}$

$V_2=\frac{P_1^{\frac{1}{\gamma }}{V}_o}{P_1^{\frac{1}{\gamma }}+P_2^{\frac{1}{\gamma }}}$

(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.

Hence heat given to the gas in the left part = 0

(c) There will be a common ressure 'P' when the equlibrium is reached.

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

For equilibrium, $V_1=V_2=\frac{V_o}{2}$

Hence,

$P_1{\left(\frac{V_o}{2}\right)}^{\gamma }+P_2{\left(\frac{V_o}{2}\right)}^{\gamma }=P(V_o{)}^{\gamma }$

or P = ${\left(\frac{P_1^{\frac{1}{\gamma }}+P_2^{\frac{1}{\gamma }}}{2}\right)}^{\gamma }$