Q42 / 180 · AIIMS 2009 · Chemistry

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

  • Correct answer
    A
    $2.303 C_P \log \left(T_2 / T_1\right)$
  • B
    $2.303 C_P \ln \left(T_2 / T_1\right)$
  • C
    $C_P \ln \left(T_1 / T_2\right)$
  • D
    $C_V \ln \left(T_2 / T_1\right)$

Solution

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 $\Delta S=2.303 C_P \log \frac{T_2}{T_1}$.

Hence, the correct answer is option: 'A'.

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