The entropy change involved in the isothermal reversible expansion of 1 mole of an ideal gas from a volume of 20 L to 200 L at 300 K is JK- 1. Give answer after rounding off to the nearest integer value.

Question

The entropy change involved in the isothermal reversible expansion of 1 mole of an ideal gas from a volume of 20 L to 200 L at 300 K is JK- 1. Give answer after rounding off to the nearest integer value. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We know that work done in isothermal reversible expansion,

q_{rev} = - w_{m a x} = 2.303 \times n RT lo g_{10} \frac{V _{2}}{V _{1}}

Given that,

n = 1, R = 8.314 J K^{- 1} m o l^{- 1}, T = 300 K

V_{1} = 20 L, V_{2} = 200 L

Putting the values in above equation.

∴ q_{rev} = 2.303 \times 1 \times 8.314 \times 300 \times (lo g)_{10} (\frac{200}{20})

= 2.303 \times 1 \times 8.314 \times 300 \times (lo g)_{10} (10)

= 2.303 \times 1 \times 8.314 \times 300

According to third law of thermodynamics,

Δ S = \frac{q _{rev}}{T}

∴ Δ S = \frac{2.303 \times 1 \times 8.314 \times 300}{300} = 19.15 J K^{- 1}

So the Entropy change is

≅ 19 J K^{- 1}

Q. The entropy change involved in the isothermal reversible expansion of 1 mole of an ideal gas from a volume of 20L to 200L at 300K is ____ JK−1. Give answer after rounding off to the nearest integer value.

Q. The entropy change involved in the isothermal reversible expansion of $1$ mole of an ideal gas from a volume of $20 L$ to $200 L$ at $300 K$ is ____ $J K^{- 1} .$
Give answer after rounding off to the nearest integer value.