A closed vessel contains 0.1 mole of a monoat | vedclass

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(D) Since the vessel is closed and insulated,no heat is exchanged with the surroundings and no work is done by or on the gas. Therefore,the total inte

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$266.67$

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(D) Since the vessel is closed and insulated,no heat is exchanged with the surroundings and no work is done by or on the gas. Therefore,the total internal energy of the system remains conserved.The internal energy of a monoatomic ideal gas is given by $U = n C_v T$,where $C_v = \frac{3}{2}R$.Let $n_1 = 0.1$ mole at $T_1 = 200\, K$ and $n_2 = 0.05$ mole at $T_2 = 400\, K$.By the principle of conservation of energy:$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_{final}$Since $C_v$ is the same for both,it cancels out:$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_{final}$$(0.1 	imes 200) + (0.05 	imes 400) = (0.1 + 0.05) T_{final}$$20 + 20 = 0.15 	imes T_{final}$$40 = 0.15 	imes T_{final}$$T_{final} = \frac{40}{0.15} = \frac{4000}{15} = \frac{800}{3} \approx 266.67\, K$.

$A$ closed vessel contains $0.1$ mole of a monoatomic ideal gas at $200\, K$. If $0.05$ mole of the same gas at $400\, K$ is added to it,the final equilibrium temperature (in $K$) of the gas in the vessel will be closest to:

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