JEE Mains · Physics · STD 11 - 11. thermodynamics
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
- A \(46.87\)
- B \(45.78\)
- C \(55.78\)
- D \(50.23\)
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 \(T _{1} V _{1}^{\gamma-1}= T _{2} V _{2}^{\gamma-1}\) \(T \cdot V ^{7 / 5-1}= T _{2}\left(\frac{ V }{10}\right)^{7 / 5-1}\)…
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