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JEE Main 2021 (25 Jul, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2021 (25 Jul, Shift 1)

Q200JEE Main 2021 · 25 JUL · SHIFT 1Difficult

Thermodynamic Processes

A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If l1 and l2 are the lengths of the gas column, before and after the expansion respectively, then the value of T1T2\frac{T_1}{T_2} will be:

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Answer: (b)

  1. Adiabatic, monatomic γ=53\gamma = \frac{5}{3}: TV2/3=TV^{2/3} = constant, and V∝lV \propto l.
  2. T1l12/3=T2l22/3T_1l_1^{2/3} = T_2l_2^{2/3}
  3. T1T2=(l2l1)2/3\frac{T_1}{T_2} = \left(\frac{l_2}{l_1}\right)^{2/3}
  4. So the answer is (b).
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