Question

By how much does the entropy of 3 mole of an ideal gas change in going from a pressure of 2 bar to a pressure of 1 bar without any change in Temperature . If the surrounding is at 1 bar and 300 K (Expansion is again of the cosntant extenal pressure of surrounding).

A+7.29J−K⁻¹
B+4.82J−K⁻¹
C−5.29J−K⁻¹
D−8.35J−K⁻¹

Answer 1

The change in entropy of the ideal gas can be determined using the entropy change equation, ΔS = R ln(V2/V1). By plugging in the given values and calculating, the change in entropy is approximately 5.29 J/K·mol.

Step-by-step explanation:

The change in entropy of an ideal gas can be determined using the equation for entropy change:

ΔS = R ln(V2/V1)

where ΔS is the change in entropy, R is the gas constant (8.314 J/K·mol), V2 is the final volume, and V1 is the initial volume.

In this case, the initial pressure (P1) is 2 bar and the final pressure (P2) is 1 bar.

Using the ideal gas law, we can find the initial and final volumes:

V1 = nRT1/P1 = 3 mol * 8.314 J/K·mol * 300 K / 2 bar = 3 * 8.314 * 300 / 2 J = 3741.9 J

V2 = nRT2/P2 = 3 mol * 8.314 J/K·mol * 300 K / 1 bar = 3 * 8.314 * 300 / 1 J = 7483.8 J

Plugging these values into the entropy change equation:

ΔS = R ln(V2/V1) = 8.314 J/K·mol * ln(7483.8 J / 3741.9 J) = 8.314 J/K·mol * ln(2)

Using a calculator, we can find that:

ΔS ≈ 5.29 J/K·mol