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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2018
  • Medium
  • MCQ

Two moles of an ideal monoatomic gas occupies a volume \(V\) at \(27^o C\). The gas expands adiabatically to a volume \(2\ V\). Calculate \((a)\) the final temperature of the gas and \((b)\) change in its internal energy.

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Answer & Solution

Correct Answer

(B) \((a)\) \(189\) \(K\)       \((b)\) \(-2.7\) \(kJ\)

Step-by-step Solution

Detailed explanation

In an adiabatic process

\(T{V^{\gamma  - 1}}=constant\)   or    \({T_1}{V_1}^{\gamma  - 1} = {T_2}{V_2}^{\gamma  - 1}\)

For monoatomic gas    \(\gamma  = \frac{5}{3}\)

\(\left( {300} \right){V^{2/3}} = {T_2}{\left( {2V} \right)^{2/3}} \Rightarrow {T_2} = \frac{{300}}{{{{\left( 2 \right)}^{2/3}}}}\)

\({T_2} = 189\,K\,\left( {final\,temperature} \right)\)

Change in internal energy \(\Delta U = n\frac{f}{2}R\,\Delta T\)

\( = 2\left( {\frac{3}{2}} \right)\left( {\frac{{25}}{3}} \right)\left( { - 111} \right) =  - 2.7\,kJ\)

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