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Heat and Thermodynamics question

2023 · 6 Apr · Shift 2 · Q47
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Heat and Thermodynamics question

2023 · 6 Apr · Shift 2 · Q47

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The temperature of an ideal gas is increased from 200 K200 \mathrm{~K}200 K to 800 K800 \mathrm{~K}800 K. If r.m.s. speed of gas at 200 K200 \mathrm{~K}200 K is v0v_{0}v0​. Then, r.m.s. speed of the gas at 800 K800 \mathrm{~K}800 K will be:
  1. A
    v0v_{0}v0​
  2. B
    2v02 v_{0}2v0​
  3. C
    4v04 v_{0}4v0​
  4. D
    v04\frac{v_{0}}{4}4v0​​
View written solutionFree

Correct answer: B

  1. For an ideal gas, the r.m.s. speed is given by vrms=3RTMv_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}vrms​=M3RT​​ Hence, vrms∝Tv_{\mathrm{rms}} \propto \sqrt{T}vrms​∝T​ when the gas remains the same.

  2. Let the r.m.s. speed at T1=200 KT_1 = 200\,\mathrm{K}T1​=200K be v1=v0v_1 = v_0v1​=v0​

  3. At T2=800 KT_2 = 800\,\mathrm{K}T2​=800K, the new r.m.s. speed is v2=v1T2T1v_2 = v_1\sqrt{\frac{T_2}{T_1}}v2​=v1​T1​T2​​​

  4. Substitute the values: v2=v0800200=v04=2v0v_2 = v_0\sqrt{\frac{800}{200}} = v_0\sqrt{4} = 2v_0v2​=v0​200800​​=v0​4​=2v0​

  5. Therefore, the correct option is: 2v0\boxed{2v_0}2v0​​ which corresponds to Option B.

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