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JEE Mains · Physics · STD 11 - 11. thermodynamics

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.

  1. A \((a)\) \(195 \) \(K\)       \((b)\) \(-2.7\) \(kJ\)
  2. B \((a)\) \(189\) \(K\)       \((b)\) \(-2.7\) \(kJ\)
  3. C \((a)\) \(195\) \(K\)      \((b)\) \(2.7\) \(kJ\)
  4. D \((a)\) \(189\) \( K\)       \((b)\) \(2.7\) \(kJ\)
Verified Solution

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 \(\gamma = \frac{5}{3}\)…
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