Question

When three moles of helium is mixed with one mole of neon at constant temperature and pressure (25 °C, 1 atm), the entropy of mixing (in J K^-1, rounded off to two decimal places) is ______

(Given: R = 8.314 JK^-1 mol^-1)

Answer 1

Concept:

• The entropy is the randomness of a system.
• When partition or separation between gases is removed, they tend to diffuse.
• This diffusion results in a change of intermolecular forces between the molecules, as a result of which there is a change in the randomness of the system. The randomness of the system increases.
• The total entropy tends to be the sum of the entropy due to individual components.
• The entropy of mixing is given by:

\(S= -R[n_A lnX_A+ n_B lnX_B ]\), where X_A and X_B are the mole fraction of individual components.

Calculation:

Given:

Number of moles of Helium = 3

• Number of moles of Neon = 1
• The total number of moles = 3 + 1 = 4
• Mole fraction of Helium = 3/4
• Mole fraction of Neon = 1/4
•  R = 8.314 JK-1 mol-1

We know, the entropy of mixing is given by:

\(S= -R[n_A lnX_A+ n_B lnX_B ]\)

• Substituting the given values in the above equation, we get:

\(Δ S = -2.303\times 8.314[3log\frac{3}{4}+1 log\frac{1}{4}]\)

or, Δ S = +18.704 J K-1

Rounding off to two decimal place we get, Δ S = +18.70 J K-1

Hence, the entropy of mixing of the two gases is 18.70 J K-1.