Question

Three moles of an ideal gas ($C_{V,m}=12.5J{K}^{-1}mo{l}^{-1}$) are at 300 K and 5 $d{m}^3$. If the gas is heated to 320 K and the volume changed to 10 $d{m}^3$, calculate the entropy change.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We know that
$\Delta S$ =$\frac{q}{T}$
$\&$ we know, from first law,
$\Delta U=q+\omega$
So, now, if we take a look at the question, temperature and volume is changing.
So, do you remember entropy change as a function of T and V?
If you don't lets derive it.
To calculate ${\Delta }_{sys}S$ for a reversible process,
As per definition: $\Delta S=\frac{d{q}_{rev}}{T}$
$\Rightarrow$ T dS = $d{q}_{rev}$
or T dS = dU + (-dw) [ From the first law of thermodynamics]
or T dS = $n{C}_{V}dT$ + P dV [dU = $n{C}_{V}dT$ and dw = -P dV]
or T ds = $n{C}_{V}dT+\frac{nRT}{V}dV$ [$P=\frac{nRT}{V}$]
$\therefore dS=\frac{n{C}_{V}dT}{T}+\frac{nR}{V}dV$
${\Delta }_{sys}S=nln\frac{T_2}{T_1}(C_V+R)+nRln\frac{P_1}{P_2}$
${\Delta }_{sys}S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$
Here,
You can see entropy change in terms of T, V and T, P. Here, we need it as a function of T and V as $C_V$ is constant.
Therefore,
$\Delta S$ =$n{C}_{V,m}ln\frac{T2}{T1}+nRln\frac{V2}{V1}$
$\Delta S$ =$\left(3mol\right)\times \left(12.5J{K}^{-1}mo{l}^{-1}\right)\times 2.303$
$log\frac{320}{300}+\left(3mol\right)\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)\times 2.303log\frac{10}{5}$
$=2.42J{K}^{-1}+17.29J{K}^{-1}=19.71J{K}^{-1}$