Question

Which of the following is the correct expression for $\Delta S$ in case of isochoric process?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

So, here what do you have?
Isochoric $\Rightarrow \Delta V=0$
Now, for an isochoric process we know that
dS = $\frac{d{q}_{rev}\left(P\right)}{T}$ Also $d{q}_{\left(rev\right)}\left(V\right)=dU$ [From first law of thermodynamics]
But dU = C-V dT
$\Rightarrow d{q}_V=C_{V}dT$
$\Rightarrow$ Integration both sides with proper limit, we get
$\Delta S=C_v[lnT{]}_{T_1}^{T_2}=C_{v}ln\frac{T_2}{t_1}$
$\Delta S=2.303C_{v}log\frac{T_2}{T_1}=C_{v}ln\frac{T_1}{t_2}$
For n moles:
$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}=2.303n{C}_{v}log\frac{T_2}{T_1}$
So, (c) is the correct option.