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JEE Main 2021 (26 Feb, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2021 (26 Feb, Shift 2)

Q197JEE Main 2021 · 26 FEB · SHIFT 2Medium

Thermodynamic Processes

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V=KT23V = KT^{\frac{2}{3}}. The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]

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

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Answer: 60

  1. V=KT2/3⇒dVV=23dTTV = KT^{2/3} \Rightarrow \frac{dV}{V} = \frac{2}{3}\frac{dT}{T}
  2. W=∫P dV=∫nRTVdV=∫nRT⋅23dTT=23nRΔTW = \int P\,dV = \int \frac{nRT}{V}dV = \int nRT \cdot \frac{2}{3}\frac{dT}{T} = \frac{2}{3}nR\Delta T
  3. For one mole: W=23×R×90=60RW = \frac{2}{3} \times R \times 90 = 60R
  4. So x=60x = 60.
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