Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. The total work done by the gases till the time they achieve equilibrium will be [2014]

Question

In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulated material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are $C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R$ and those for an ideal diatomic gas are $C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R .$

Q. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. The total work done by the gases till the time they achieve equilibrium will be [2014]

Smart Achievers · Accepted Answer

(4)

$As the pressure of gases in both compartments is the same$

$∴$ $n C_{p 1} (700 - T) = n C_{p 2} (T - 400)$

$\frac{5}{2} R (700 - T) = \frac{7}{2} R (T - 400)$

$3500 - 5 T = 7 T - 2800$ $\Rightarrow 12 T = 6300$

$∴$ $T = 525 K$

Applying first law of thermodynamics,

$Δ W_{1} + Δ U_{1} = Δ Q_{1}$

and $Δ W_{2} + Δ U_{2} = Δ Q_{2}$

$Δ Q_{1} + Δ Q_{2} = 0$

or, $- (Δ W_{1} + Δ W_{2}) = Δ U_{1} + Δ U_{2}$

$= n_{1} C_{v 1} (525 - 700) + n_{2} C_{v 2} (525 - 400)$

$= - 2 \times \frac{3 R}{2} \times 175 + 2 \times \frac{5 R}{2} \times 125$

$= - 525 R + 625 R = - 100 R$

Therefore, total work done $= - 100 R$

Solution

1 250 R

2 200 R

3 100 R

4 −100 R

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