Question

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

Short answer

The entropy change (ΔS) for the isothermal process where the volume of helium gas is increased from 2.00 L to 5.00 L at 300 K, is approximately \(2.071~J/K\).

Step-by-step solution

Convert temperature to Kelvin

In thermodynamics, it's crucial to use the Kelvin temperature scale. To convert the given temperature (27°C) to Kelvin, we need to add 273: T = 27°C + 273 = \(300 K\) Now, we have the temperature in Kelvin and are ready to proceed.

Use the formula for entropy change

We will use the formula for entropy change in an isothermal process for an ideal gas: ΔS = nR * ln(V2/V1) We have: - n = 0.100 mol - R = 8.314 J/mol K (gas constant) - V1 = 2.00 L - V2 = 5.00 L

Calculate the entropy change

Plug the values into the ΔS formula and calculate the entropy change: ΔS = (0.100 mol) * (8.314 J/mol K) * ln(\(5.00 L / 2.00 L\)) ΔS ≈ (0.100 mol) * (8.314 J/mol K) * ln(2.5) ΔS ≈ \(2.071~J/K\) The entropy change for the process is approximately \(2.071~J/K\).