Question

In an isobaric process, when temperature changes from $T_1$ to $T_2$, $\Delta S$ is equal to

• A. $2.303\, C_{P} \log \left(T_{2} / T_{1}\right)$
• B. $2.303 \,C_{P} \log \left(T_{2} / T_{1}\right)$
• C. $C_{p} \ln \left(T_{1} / T_{2}\right)$
• D. $C_{V} \ln \left(T_{2} / T_{1}\right)$

Answer 1

Answer: (A)

The entropy change for a process, when $T$ and $P$ are the variables is given by

$\Delta S=C_{P} \,In\, \frac{T_{2}}{T_{1}}-R\, In\, \frac{P_{2}}{P_{1}}$

For an isobaric process $P_1 = P_2$.

Hence the above equation reduces to

$C_{P}In \frac{T_{2}}{T_{1}}=\Delta S.$

or $\Delta S=2.303C_{P}\, log \frac{T_{2}}{T_{1}}.$