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A cylinder of volume $V = 10^{- 2} m^{3}$ made by non conducting walls contains $8 gm$ of He gas. The cylinder is divided by a thin fixed membrane into two unequal parts. Volume of the left compartment is $V / 3$ and the right component has volume $\frac{2 V}{3}$ .
The left compartment has an installed electric heater. Due to finite conductivity of the membrane, heat passes from the left part into the right part. Heat transfer rate through the membrane is 0.2 watt per $ΔT = 1^{\circ} C$ temperature difference across the membrane. Initially the heater is off and both the parts are in thermal and mechanical equilibrium. The membrane can withstand maximum pressure difference of $10^{3} Pa$ . The heater starts supplying heat slowly the left compartment at its maximum possible power such that the membrane remains intact (after long time of heating also). (Take $R = 25 / 3 J / mol - K$ ).

Column-I

Column-II
(I)
For the left compartment
(p)
(II)
For the right compartment
(q)
Rate of transfer of heat is 0.20 watt
(III)
For the whole cylinder
(r)
Rate of transfer of the heat is 0.06 watt
(IV)
Through the membrane
(s)
Rate of transfer of heat is 0.12 watt
The correct option is:

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(I)-(S) (II)-(R) (III)-(P) (IV)-(S)

(I)-(R) (II)-(S) (III)-(P) (IV)-(S)

(I)-(S) (II)-(R) (III)-(S) (IV)-(P)

(I)-(R) (II)-(S) (III)-(S) (IV)-(P)

answer is B.

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Detailed Solution

For left chamber : $P_{1} \frac{V}{3} = \frac{2}{3} R T_{1}$
For right chamber : $P_{2} \frac{2 V}{3} = \frac{4}{3} R T_{2}$
$\Rightarrow P_{1} - P_{2} = 2 \frac{R}{V} (T_{1} - T_{2})$
$\Rightarrow T_{1} - T_{2} = \frac{10^{3} \times 10^{- 2}}{2 \times \frac{25}{3}} = 0.6$ kelvin = 0.6°C
Hence
Rate of the conduction through membrane = rate of heat in 2nd compartment
$= 0.2 \times 0.6 = 0.12 W$
Hence II – S, IV – S
Also, P = Power of heater
$\frac{4}{3} \times \frac{3 R}{2} \frac{d T_{2}}{d t} = 0.12 \frac{2}{3} \times \frac{3 R}{2} \frac{d T_{1}}{d t} = P - 0.12 \Rightarrow R (\frac{d T_{1}}{d t} - \frac{d T_{2}}{d t}) = P - 0.12 - 0.06 = P - 0.18 \Rightarrow P - 0.18 = R \frac{d}{d t} (T_{1} - T_{2}) = 0$

As $T_{1} - T_{2} =$ constant at maxm pressure difference.

$\Rightarrow P = 0.18 W$

Therefore: Rate of heat in 1st compartment

Hence: I – R
For whole system let
P = average pressure, T = average temperature
$∵$ V = constant
$\Rightarrow \frac{P}{T} =$ constant
Hence III – P

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