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JEE Main 2014 (9 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2014 (9 Apr, Shift 1)

Q232JEE Main 2014 · 9 APR · SHIFT 1Difficult

Thermodynamic Processes

The equation of state for a gas is given by PV = nRT + α\alphaV, where n is the number of moles and α\alpha is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are To and Po respectively. The work done by the gas when its temperature doubles isobarically will be:

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Answer: (a)

  1. For 1 mole: PV=RT+αV⇒V=RTP−αPV = RT + \alpha V \Rightarrow V = \frac{RT}{P - \alpha}.
  2. At constant PoP_o, To→2ToT_o \to 2T_o: ΔV=RToPo−α\Delta V = \frac{RT_o}{P_o - \alpha}
  3. W=PoΔV=PoToRPo−αW = P_o\Delta V = \frac{P_oT_oR}{P_o - \alpha}
  4. So the answer is (a).
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