AP EAMCET · PHYSICS · Thermodynamics
2 moles of a monatomic gas requires heat energy \(Q\) to be heated from \(30^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) at constant volume. The heat energy required to raise the temperature of 4 moles of a diatomic gas from \(28^{\circ} \mathrm{C}\) to \(33^{\circ} \mathrm{C}\) at constant volume is
- A \(2 \mathrm{Q}\)
- B \(\frac{7 Q}{2}\)
- C \(\frac{4 Q}{3}\)
- D \(\frac{5 Q}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{5 Q}{3}\)
Step-by-step Solution
Detailed explanation
Heat energy, \(\mathrm{Q}\) at constant volume, \(\mathrm{Q}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}\) for monatomic gas \(\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\) and for diatomic gas, \(\mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}\)…
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