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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

  1. A \(46.87\)
  2. B \(45.78\)
  3. C \(55.78\)
  4. D \(50.23\)
Verified Solution

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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