Question

Which of the following is the correct expression of $\Delta S$ for an isobaric process?

• A. $\Delta S=n{C}_{P}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
• B. $\Delta S=n{C}_{p}ln\frac{T_2}{T_1}$
• C. $\Delta S=n{C}_{p}ln\frac{T_1}{T_2}$
• D. None of these

Answer 1

The correct option is C

$\Delta S=n{C}_{p}ln\frac{T_1}{T_2}$

So, what is the first thing which comes to our mind?
Yes, Isobaric $\Rightarrow$ constant pressure
Now, as we have already seen that
Entropy change in an isobaric process (at constant pressure)
dS = $\frac{d{q}_{rev\left(P\right)}}{T}$ also $d{q}_{rev\left(P\right)}$ = dH
But $dH=C_{P}dT$, substituting for relationship with entropy,
We get,
$\Rightarrow dS=C_P\frac{dT}{T}$
Integrating both sides
${\int }_{S1}^{S2}ds=C_P{\int }_{T1}^{T2}\frac{dT}{T}$
$\Delta S=C_p[lnT{]}_{T_1}^{T_2}=C_{p}ln\frac{T_2}{T_1}$
$\Delta S=2.303C_{p}log\frac{T_2}{T_1}=\Delta S=C_{p}ln\frac{T_2}{T_1}$
For n moles:
$\Delta S=n{C}_{P}ln\frac{T_2}{T_1}=2.303n{C}_{P}log\frac{T_2}{T_1}$