Question

1 mole of an ideal gas at 25⁰C is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in J k–1mol–1)

• A. 19.15
• B. – 19.15
• C. 4.7
• D. Zero

Answer 1

Answer: (D)

Zero

Answer 2

$\Delta$S = nCv $\mathcal{l}$n $\frac{T_{2}}{T_{1}}$ + nR $\mathcal{l}$n $\frac{V_{2}}{V_{1}}$

For adiabatic process (Q = 0)

$\Delta$E = W $\Rightarrow$ nCv $\mathcal{l}$n $\frac{T_{2}}{T_{1}}$= – nR $\mathcal{l}$n $\frac{V_{2}}{V_{1}}$

$\therefore$ $\Delta$S = 0.