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Thermodynamics question

2020 · 4 Sep · Shift 2 · Q18
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Thermodynamics question

2020 · 4 Sep · Shift 2 · Q18

JEE MainChemistryThermodynamicsMCQ+4 / −1
Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :
  1. A
    Zero
  2. B
    -RT ln⁡V2V1\ln {{{V_2}} \over {{V_1}}}lnV1​V2​​
  3. C
    CV (T2 – T1)
  4. D
    – RT (V2 – V1)
View written solutionFree

Correct answer: A

  1. Identify the process

The gas is expanded into vacuum. This is called free expansion.

In free expansion:

  • External pressure opposing expansion is zero.
  • Hence, work done is calculated by

w=−∫Pext dVw = -\int P_{\text{ext}}\, dVw=−∫Pext​dV

  1. Substitute for expansion into vacuum

Since the gas expands into vacuum,

Pext=0P_{\text{ext}} = 0Pext​=0

Therefore,

w=−∫0 dV=0w = -\int 0\, dV = 0w=−∫0dV=0

  1. Conclusion

So, irrespective of the number of moles and the doubling of volume, the work done in free expansion is

w=0w = 0w=0

  1. Check options
  • A: Zero — Correct
  • B: −RTln⁡V2V1-RT\ln \dfrac{V_2}{V_1}−RTlnV1​V2​​ — This is the work expression for reversible isothermal expansion of 1 mole, not free expansion.
  • C: CV(T2−T1)C_V(T_2-T_1)CV​(T2​−T1​) — This corresponds to change in internal energy for an ideal gas, not work.
  • D: −RT(V2−V1)-RT(V_2-V_1)−RT(V2​−V1​) — Incorrect expression for work.

Hence, the correct option is A.

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