JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
\(1\,mole\) of ideal gas is allowed to expand reversibly and adiabatically from a temperature of \(27^{\circ}\,C\). The work done is \(3\,kJ\,mol ^{-1}\). The final temperature of the gas is \(......K\) (Nearest integer). Given \(C _{ V }=20\,J\,mol ^{-1}\,K ^{-1}\)
- A \(120\)
- B \(130\)
- C \(140\)
- D \(150\)
Answer & Solution
Correct Answer
(D) \(150\)
Step-by-step Solution
Detailed explanation
\(q =0\) \(\Delta U = w\) \(1 \times 20 \times\left[ T _2-300\right]=-3000\) \(T _2-300=-150\) \(T _2=150\,K\)
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