Question

One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25⁰C and 1 L to the state when temperature is 100⁰C and volume 10 L. The entropy change of the process can be expressed as (R = 2 calories/mol/K)

• A. 3 ln$\frac{298}{373}$+ 2 ln 10
• B. 5 ln$\frac{373}{298}$ + 2 ln 10
• C. 7 ln$\frac{373}{298}$+ 2 ln$\frac{1}{10}$
• D. 5 ln$\frac{373}{298}$+ 2 ln $\frac{1}{10}$

Answer 1

Answer: (B)

5 ln$\frac{373}{298}$ + 2 ln 10

Answer 2

$\Delta$S = nCV ln $\left( \frac{T_{f}}{T_{i}} \right)$ + nR ln $\left( \frac{V_{f}}{V_{i}} \right)$ = $\frac{5}{2}$×2 ln $\frac{373}{298}$+ 2 ln 10