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JEE Main 2019 (10 Apr, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2019 (10 Apr, Shift 2)

Q218JEE Main 2019 · 10 APR · SHIFT 2Difficult

Thermodynamic Processes

One mole of ideal gas passes through a process where pressure and volume obey the relation {}

P=P0[1−12(V0V)2]P = P_0\left[1 - \frac{1}{2}\left(\frac{V_0}{V}\right)^{2}\right]. Here P0 and V0 are constants. Calculate the change in the temperature of the gas if its volume changes form V0 to 2V0

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Answer: (c)

  1. One mole: T=PVRT = \frac{PV}{R}.
  2. At V0V_0: P=P02P = \frac{P_0}{2}, so T1=P0V02RT_1 = \frac{P_0V_0}{2R}.
  3. At 2V02V_0: P=P0(1−18)=7P08P = P_0\left(1 - \frac{1}{8}\right) = \frac{7P_0}{8}, so T2=7P0V04RT_2 = \frac{7P_0V_0}{4R}.
  4. ΔT=74−24=54P0V0R\Delta T = \frac{7}{4} - \frac{2}{4} = \frac{5}{4}\frac{P_0V_0}{R}, so the answer is (c).
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