Question

Calculate the entropy of a mixture of 50% Ne and 50% Ar at 500 K and 10 atm, assuming ideal behavior.

Answer 1

Answer:

5.76 J/K

Explanation:

Mole fraction of Ne at 500 K= 50 % = 0.5 = $x_A$

Mole fraction of Ar at 500 K = 50 %= 0.5 = $x_B$

R = Gas constant = 8.314 J/Kmol

As mass is not given the number of moles of Ne and Ar are taken as 0.5

Entropy of mixture

$\Delta_{mix}S=-R(n_Alnx_A+n_Blnx_B)\\\Rightarrow \Delta_{mix}S=-8.314(0.5ln0.5+0.5ln0.5)\\\Rightarrow \Delta_{mix}S=5.76\ J/K$

∴ Entropy of mixture is 5.76 J/K