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In an isobaric process, when temperature changes from $T_1$ to $T_2, \Delta S$ is equal to
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Verified Answer
The correct answer is:
$2.303 C_P \log \left(T_2 / T_1\right)$
The entropy change for a process, when $T$ and $P$ are the variables is given by
$\Delta S=C_P \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}$
For an isobaric process $P_1=P_2$. Hence the above equation reduces to
$C_P \ln \frac{T_2}{T_1}=\Delta S$
or $\Delta S=2.303 C_P \log \frac{T_2}{T_1}$.
$\Delta S=C_P \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}$
For an isobaric process $P_1=P_2$. Hence the above equation reduces to
$C_P \ln \frac{T_2}{T_1}=\Delta S$
or $\Delta S=2.303 C_P \log \frac{T_2}{T_1}$.
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