What entropy change is involved in the isothermal expansion of 5 mol of ideal gas from a volume of 10 L to 100 L at 300 K?

Question

What entropy change is involved in the isothermal expansion of 5 mol of ideal gas from a volume of 10 L to 100 L at 300 K? (A) -95.7 J K -1 (B) +95.7 J K -1 (C) 28.72 kJ K -1 (D) -28.72 kJ K -1

Tardigrade Admin · Accepted Answer

Δ S=n R ln (V2/V1)=2.303 n R log (V2/V1) =2.303 × 5 mol ×(8.314 JK -1 mol -1) log (100 L /10 L )=95.7 JK -1

Q. What entropy change is involved in the isothermal expansion of $5 m o l$ of ideal gas from a volume of $10 L$ to $100 L$ at $300 K$ ?

$- 95.7 J K^{- 1}$

$+ 95.7 J K^{- 1}$

$28.72 k J K^{- 1}$

$- 28.72 k J K^{- 1}$

$Δ S = n R ln \frac{V _{2}}{V _{1}} = 2.303 n R lo g \frac{V _{2}}{V _{1}}$

$= 2.303 \times 5 m o l \times (8.314 J K^{- 1} m o l^{- 1}) lo g \frac{100 L}{10 L} = 95.7 J K^{- 1}$

Q. What entropy change is involved in the isothermal expansion of 5mol of ideal gas from a volume of 10L to 100L at 300K?

−95.7JK−1

+95.7JK−1

28.72kJK−1

−28.72kJK−1