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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2021 · 25 July · Morning Shift · Physics · Thermodynamics

A monoatomic ideal gas, initially at temperature T _ 1 is enclosed...

A monoatomic ideal gas, initially at temperature T1{T}_{1} is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2{T}_{2} by releasing the piston suddenly. If l1{l}_{1} and l2{l}_{2} are the lengths of the gas column, before and after the expansion respectively, then the value of T1T2\dfrac{{T}_{1}}{{T}_{2}} will be:

Options

  1. A(l1l2)23\left(\dfrac{{l}_{1}}{{l}_{2}}\right)\dfrac{2}{3}
  2. B(l2l1)23\left(\dfrac{{l}_{2}}{{l}_{1}}\right)\dfrac{2}{3}
  3. C(l2l1)\left(\dfrac{{l}_{2}}{{l}_{1}}\right)
  4. Dl1l2\dfrac{{l}_{1}}{{l}_{2}}

Correct answer

B. (l2l1)23\left(\dfrac{{l}_{2}}{{l}_{1}}\right)\dfrac{2}{3}

Step-by-step solution

PVr=P{V}^{r} = constant

TVr−1=T{V}^{r - 1} = constant

T(ℓ)53−1=T{( ℓ )}^{\dfrac{5}{3} - 1} = constant

T1T2=(ℓ2ℓ1)2/3\dfrac{{T}_{1}}{{T}_{2}} = {\left(\dfrac{{ℓ}_{2}}{{ℓ}_{1}}\right)}^{2 / 3}

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