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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

  1. Same monoatomic gas, closed vessel, no energy lost: (n1+n2)T=n1T1+n2T2(n_1 + n_2)T = n_1T_1 + n_2T_2.
  2. T=0.1×200+0.05×4000.15=400.15T = \frac{0.1 \times 200 + 0.05 \times 400}{0.15} = \frac{40}{0.15}
  3. T≈266.7T \approx 266.7 K, so about 266266 K.
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