A container has 0.2 mole of O₂ gas and the gas is heated from 0 $\circ$ C to 100 $\circ$ C.What is the change in entropy of the gas? (R = 8.314 J/mol·K)

Question

Quizplus · Accepted Answer

First, we need to calculate the initial and final entropy of the gas using the formula: ΔS = nCln(T2/T1) - nRln(V2/V1) where ΔS is the change in entropy, n is the number of moles, C is the molar heat capacity at constant volume, R is the gas constant, T1 and T2 are the initial and final temperatures, and V1 and V2 are the initial and final volumes (assuming constant volume). At constant volume: ΔS = nCvln(T2/T1) - nRln(V2/V1) where Cv is the molar heat capacity at constant volume. Since O₂ gas is a diatomic gas, Cv = (5/2)R. Plugging in the values, we get: ΔS = (0.2 mol)(5/2)(8.314 J/mol·K)ln(100/0) - (0.2 mol)(8.314 J/mol·K)ln(1) ΔS = (2.61 J/K/mol)(ln10) ΔS ≈ 1.82 J/K Therefore, the correct answer is A) 1.82 J/K.

A Container Has 0 ∘\circ∘ C to 100 ∘\circ∘ CWhat Is the Change in Entropy of the Gas? (R

A Container Has 0 $\circ$ C to 100^$\circ$CWhat Is the Change in Entropy of the Gas? (R