AP EAMCET · PHYSICS · Thermodynamics
The work done by 6 moles of helium gas when its temperature increases by \(20^{\circ} \mathrm{C}\) at constant pressure is
(Universal gas constant \(=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) )
- A 807.2 J
- B 887.2 J
- C 997.2 J
- D 1007.2 J
Answer & Solution
Correct Answer
(C) 997.2 J
Step-by-step Solution
Detailed explanation
\(W = nR\Delta T\) \(W = 6 \times 8.31 \times 20\) \(W = 997.2 \mathrm{~J}\)
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