Question

The figure shows an insulated cylinder divided into three parts

A, B

and

C

. Pistons

I

and

II

are connected by a rigid rod and can move without friction inside the cylinder. Piston

I

is perfectly conducting while piston

II

is perfectly insulating. The initial state of the gas

(γ = 1.5)

present in each compartment

A, B

and

C

is as shown. Now, compartment,

A

is slowly given heat through a heater,

H

such that the final volume of

C

becomes

\frac{4 V_{0}}{9}

. Assume the gas to be ideal and find.

(a) Final pressures in each compartment $A, B$ and $C$

(b) Final temperatures in each compartment $A, B$ and $C$

(c) Heat supplied by the heater

(d) Work done by gas in $A$ and $B$ .

(e) Heat flowing across piston $I$ .

Accepted Answer

(a) For compartment - $C$ , from the idea of adiabatic process,

$P V^{γ} =$ constant

$\Rightarrow P_{0} V_{0}^{\frac{3}{2}} = P_{1} {(\frac{4 V_{0}}{9})}^{\frac{3}{2}}$

$\Rightarrow P_{1} = P_{0} {(\frac{9}{4})}^{\frac{3}{2}} = \frac{27 P_{0}}{8}$

$P_{1} =$ Final pressure in $A =$ final pressure in $C$ , as movable piston is in equilibrium.

Temperature in compartment $' A'$

$\Rightarrow \frac{P_{0} V_{0}}{T_{0}} = \frac{P_{1} (\frac{14 V_{0}}{9})}{T_{0}}$

$\Rightarrow \frac{P_{0} V_{0}}{T_{0}} = (\frac{27}{8}) \frac{P_{0} \times (\frac{14}{9}) V_{0}}{T}$

$\Rightarrow T = \frac{21}{4} T_{0}$

Temperature in $A =$ Temperature in $B$ (as conducting)

for $C$ ,

$\frac{P_{0} V_{0}}{T_{0}} = \frac{4 V_{0}}{9} \times \frac{27 P_{0}}{8 T_{C}}$

$\Rightarrow T_{C} = 4 \times \frac{27}{9 \times 8} T_{0} = \frac{3}{2} T_{0}$

pressure in $' B' = P_{B}$

Then,

$\frac{P_{B}}{T} = \frac{P_{0}}{T_{0}}$

$\Rightarrow P_{B} = \frac{P_{0} T}{T_{0}} = \frac{21}{4} \frac{T_{0} P_{0}}{T_{0}} = \frac{21}{4} P_{0}$

Heat supplied by the heater $= Q_{n}$

$\Rightarrow Q_{H} = Δ Q_{a} + Δ Q_{b}$

$\Rightarrow Δ Q_{b} = n C_{V} Δ T = \frac{f}{2} (n R T_{2} - n R T_{1})$

$\Rightarrow Δ Q_{b} = \frac{f}{2} (V_{0}) (P_{B} - P_{0})$

$\Rightarrow Δ Q_{b} = \frac{f}{2} V_{0} (\frac{17}{4} P_{0})$

$\Rightarrow Δ Q_{b} = \frac{17 f P_{0} V_{0}}{8}$

$∵ γ = 1.5 \Rightarrow f = 4$

$\Rightarrow Δ Q_{b} = \frac{17}{2} P_{0} V_{0}$

For $A$ ,

$P_{A} = P_{C}$

$\Rightarrow d V_{A} = - d V_{C}$

$\Rightarrow W_{A} = \int P_{A} d V_{A} - \int P_{C} d V_{C} = - W_{c}$

$\Rightarrow W_{A} = \frac{\frac{4 V_{0}}{9} \times \frac{27 P_{0}}{8} - P_{0} V_{0}}{\frac{1}{2}}$

$\Rightarrow W_{A} = \frac{P_{0} V_{0}}{\frac{1}{2}} = P_{0} V_{0}$

So change in internal energy at the compartment, $A$ ,

$\begin{array}{l} Δ U_{A} = \frac{f}{2} (n R Δ T_{A}) = (P_{1} \times \frac{14 V_{0}}{9} - P_{0} V_{0}) \frac{f}{2} \end{array}$

$\Rightarrow Δ U_{A} = (\frac{27}{8} P_{0} \times \frac{14 V_{0}}{9} - P_{0} V_{0}) \frac{f}{2}$

$Δ U_{A} = (\frac{21}{4} P_{0} V_{0} - P_{0} V_{0}) \frac{f}{2}$

$Δ U_{A} = (\frac{17}{4} P_{0} V_{0}) \times 2$

$Δ Q = \frac{17}{4} P_{0} V_{0} + P_{0} V_{0} + \frac{17}{2} P_{0} V_{0}$

$\Rightarrow ∆ Q = 18 P_{0} V_{0}$

$\Rightarrow W_{a} = P_{0} V_{0}$

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