One mole of an ideal diatomic gas (C v = 5 cal) was transformed from initial 25°C and 1 L…
Chemistry · JEE Main · NTA Exams — Chemical Thermodynamics
One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25°C and 1 L to the state when temperature is 100°C and volume 10 L. The entropy change of the process can be expressed as (R = 2 calories/mol/K)
- \(3 \ell \mathrm{n} \frac{298}{373}+2 \ell \mathrm{n} 10\)
- \(5 \ell \mathrm{n} \frac{373}{298}+2 \ell \mathrm{n} 10\)
- \(7 \ell \mathrm{n} \frac{373}{298}+2 \ell \mathrm{n} \frac{1}{10}\)
- \(5 \ell \mathrm{n} \frac{373}{298}+2 \ell \mathrm{n} \frac{1}{10}\)
Answer
(B) 5 n 373 298 +2 n 10
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