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JEE Main 2020 (4 Sep, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2020 (4 Sep, Shift 1)

Q31JEE Main 2020 · 4 SEP · SHIFT 1Medium

Degrees of Freedom & Internal Energy

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 close to _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 266

Same monoatomic gas, closed vessel, no energy lost: $(n_1 + n_2)T = n_1T_1 + n_2T_2$ .
$T = \frac{0.1 \times 200 + 0.05 \times 400}{0.15} = \frac{40}{0.15}$
$T \approx 266.7$ K, so about $266$ K.

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