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First Law of Thermodynamics

The figure sh...

Question

The figure shows an insulated cylinder divided into three parts A, B and C. Piston 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 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 the heat supplied by the heater.

18 P_{o} V_{o}

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12 P_{o} V_{o}

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9 P_{o} V_{o}

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25 P_{o} V_{o}

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Solution

The correct option is A $18 P_{o} V_{o}$
heat supplied $Δ Q = Δ U + Δ W$
$Δ U = Δ U_{A} + Δ U_{B} + 0$ [no heat change in case of compartment C]
$Δ W = Δ W_{A} + 0 + Δ W_{C}$ [no change in volume in case of compartment B hence work done = 0]

initial conditions : $P_{o}$ Pressure $V_{o}$ Volume $T_{o}$ Temperature
final conditions :

compartment C

$P V^{γ} = P V^{γ} = P \times (\frac{4 V_{0}}{9})^{1.5} = P_{o} V_{o}^{1.5}$
$P = \frac{27 P_{o}}{8}$
$\frac{P_{o} V_{o}}{T_{o}} = \frac{\frac{27 P_{o}}{8} \times \frac{4 V_{o}}{9}}{T}$
$T = \frac{3 T_{o}}{2}$

compartment A

$P = \frac{27 P_{o}}{8}$ compartment A and C has to have same pressure for pistons to come at rest.
$\frac{P_{o} V_{o}}{T_{o}} = \frac{\frac{27 P_{o}}{8} \times (V_{o} + \frac{5 V_{o}}{9})}{T}$
$T = \frac{21 T_{o}}{4}$

compartment B

$T = \frac{21 T_{o}}{4}$ temperature of compartment A and B should be same at equilibrium
$\frac{P_{o} V_{o}}{T_{o}} = \frac{P \times V_{o}}{\frac{21 T_{o}}{4}}$
$P = \frac{21 P_{o}}{4}$

$γ = \frac{C_{p}}{C_{v}} = \frac{f + 2}{f}$
$f = 4$
$C_{v} = \frac{f R}{2} = 2 R$

$Δ U_{A} = Δ U_{B}$
$Δ U_{A} = n C_{v} Δ T$

$Δ U_{A} = \frac{P_{0} V_{o}}{R T_{o}} \times C_{v} (\frac{21 T_{o}}{4} - T_{o}) = \frac{P_{o} V_{o}}{R T_{o}} \times 2 R \times \frac{17 T_{o}}{4} = \frac{17 P_{o} V_{o}}{2}$

$Δ W_{A} = - Δ W_{C}$ as the gas in chamber A is working on chamber C
$Δ Q_{C} = 0$ as its a Adibatic process hence
$Δ Q_{C} = Δ U_{C} + Δ W_{C}$
$Δ U_{C} = - Δ W_{C} = n C_{v} Δ T = \frac{P_{o} V_{o}}{R T_{o}} \times 2 R \times (\frac{3 T_{o}}{2} - T_{o}) = P_{o} V_{o}$

$Δ U = Δ U_{A} + Δ U_{B} + 0$ [no heat change in case of compartment C]
$Δ W = Δ W_{A} + 0 + Δ W_{C}$ [no change in volume in case of compartment B hence work done = 0]

head supplied by the heater = heat supplied to compartment A + heat flown through piston I
$Δ Q = Δ U_{A} + Δ U_{B} + Δ W_{A B} = Δ U_{A} + Δ U_{B} + Δ W_{A} = 2 Δ U_{A} + Δ W_{A}$
$Δ Q = 18 P_{o} V_{o}$

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