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.
Pick an option to check your answer.
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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