In adiabatic container with adiabatic piston contains $4 g m$ of $N_{2}$ gas and $0.8 g m$ of $H e$ gas are separated by a weakly conducting light partition. Both piston and partition are free to move. Initial temperature of $N_{2}$ and $H e$ gases are $3 T_{0} a n d T_{0}$ respectively. Initially system is at rest and in equilibrium. Final temperature of the gas is

2 T_{0}

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\frac{3 T_{0}}{2}

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\frac{7 T_{0}}{4}

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\frac{5 T_{0}}{2}

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Solution

The correct option is A $2 T_{0}$
As the system of both the gases is adiabatic, the heat cannot escape from the system.
The temperature of $N_{2}$ is more than $H e$ , so there will be flow of heat from $N_{2}$ to $H e$ to make the temperature of both gases at same value.

So, we can say that:
Heat lost by $N_{2} =$ Heat gained by $H e$
$(n C_{p} Δ T)_{N_{2}} = (n C_{p} Δ T)_{H e} . . . . . (1)$

We know that,
$C_{p} = R + C_{v} = R + \frac{f R}{2}$

So, for $N_{2}$ , $C_{p} = R + \frac{5 R}{2} = \frac{7 R}{2}$

and for $H e$ , $C_{p} = R + \frac{3 R}{2} = \frac{5 R}{2}$

Therefore, from $(1)$ , and assuming $T$ being the common temperature of both the gases:
$\frac{4}{28} (\frac{7 R}{2}) (3 T_{0} - T) = \frac{0.8}{4} (\frac{5 R}{2}) (T - T_{0})$

$\Rightarrow 2 T = 4 T_{0}$
$\Rightarrow T = 2 T_{0}$

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