JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2020
- Hard
- MCQ
An engine takes in \(5\) moles of air at \(20\,^{\circ} C\) and \(1\) \(atm,\) and compresses it adiabaticaly to \(1 / 10^{\text {th }}\) of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be \(X\, kJ\). The value of \(X\) to the nearest integer is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(45.78\)
Step-by-step Solution
Detailed explanation
Diatomic :
\(f=5\)
\(\gamma=7 / 5\)
\(T _{ i }= T =273+2 0 =293 K\)
\(V_{i}=V\)
\(V _{ f }= V / 10\)
Adiabatic \(TV ^{\gamma-1}=\) constant
\(T _{1} V _{1}^{\gamma-1}= T _{2} V _{2}^{\gamma-1}\)
\(T \cdot V ^{7 / 5-1}= T _{2}\left(\frac{ V }{10}\right)^{7 / 5-1}\)
\(\Rightarrow T _{2}= T .10^{2 / 5}\)
\(\Delta U =\frac{ nfR \left( T _{2}- T _{1}\right)}{2}=\frac{5 \times 5 \times \frac{25}{3} \times\left( T \cdot 10^{2 / 5}- T \right)}{2}\)
\(=\frac{25 \times 25 \times}{6} T \left(10^{2 / 5}-1\right)\)
\(=\frac{625 \times 293 \times\left(10^{2 / 5}-1\right)}{6}\)
\(=4.033 \times 10^{3} \approx 4 kJ\)
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