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JEE Main 2013 (25 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2013 (25 Apr, Shift 1)

Q235JEE Main 2013 · 25 APR · SHIFT 1Medium

Thermodynamic Processes

In the isothermal expansion of 10 g of gas from volume V to 2V the work done by the gas is 575 J. What is the root mean square speed of the molecules of the gas at that temperature?

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Answer: (c)

  1. Isothermal work: W=mMRTln⁡2⇒RTM=Wmln⁡2W = \frac{m}{M}RT\ln 2 \Rightarrow \frac{RT}{M} = \frac{W}{m\ln 2}
  2. vrms=3RTM=3Wmln⁡2=3×5750.01×0.693v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3W}{m\ln 2}} = \sqrt{\frac{3 \times 575}{0.01 \times 0.693}}
  3. vrms=2.49×105≈499 m/sv_{rms} = \sqrt{2.49 \times 10^5} \approx 499\text{ m/s}
  4. So the answer is (c) 499 m/s.
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