Question

If $1$mol ideal gas expanded from $5\mathrm{L}$ to $50\mathrm{L}$ at ${27}^{°}\mathrm{C}$ then entropy change in the process will be?

Answer 1

Step 1: Given data

Number of a mole of an ideal gas$=1$

Initial volume$=5\mathrm{L}$

Final volume$=50\mathrm{L}$

Temperature$={25}^{\circ }\mathrm{C}$

Step 2:Calculating entropy

$∆\mathrm{S}=2.303\mathrm{nR}\log \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$ where $\mathrm{n}=$mole of the gas, $\mathrm{R}=$gas constant, ${\mathrm{V}}_2=$final volume, ${\mathrm{V}}_1=$Initial volume$$\begin{gathered} ∆\mathrm{S}=2.303\times 1\times 8.314\log \left(\frac{50}{5}\right) \\ ∆\mathrm{S}=2.303\times 8.314\log 10 \\ ∆\mathrm{S}=2.303\times 8.314\times 1 \\ ∆\mathrm{S}=19.14\mathrm{J}{\mathrm{K}}^{-1} \end{gathered}$$

Therefore, the entropy of the process $=19.14\mathrm{J}{\mathrm{K}}^{-1}$