TS EAMCET · Physics · Thermodynamics
An office room contains about 2000 moles of air. The change in the internal energy of this much air when it is cooled from \(34^{\circ} \mathrm{C}\) to \(24^{\circ} \mathrm{C}\) at a constant pressure of \(1.0 \mathrm{~atm}\) is [Use \(\gamma_{\text {air }}=1.4\) and whiversal gas constant \(=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}]\)
- A \(-1.9 \times 10^5 \mathrm{~J}\)
- B \(+1.9 \times 10^5 \mathrm{~J}\)
- C \(-4.2 \times 10^5 \mathrm{~J}\)
- D \(+0.7 \times 10^5 \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(-4.2 \times 10^5 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Given, Number of moles of air in room \((n)=2000=2 \times 10^3\) Temperature difference \((\mathrm{dT})=24-34=-10^{\circ} \mathrm{C}\) We know that,…
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