Entropy of a solution question( please help)?
At 25 degrees celcius and 1 atm pressure, the absolute third law entropies of methane and ethane are 186.19 J/(K*mol) and 229.49 J/(K*mol) respectively in the gas phase. Calculate the absolute third-law entropy of a solution containing 1 mole of each gas. Assume Ideal behaviour.
- hfshawLv 71 decade agoFavorite Answer
The absolute entropy of an ideal solution (which can be a gas, liquid, or solid) of two components is given by:
S = (n_a)*s_a + (n_b)s_b + ΔS_mixing
where n_a is the number of moles of component a, n_b is the number of moles of component b, s_a and s_b are the absolute molar entropies of components a and b, respectively, and ΔS_mixing is the ideal entropy of mixing:
ΔS_mixing = -(n_a + n_b)*R*[x_a*ln(x_a) + x_b*ln(x_b)]
where x_a and x_b are the mole fractions of components a and b in the mixture. x_a = n_a/(n_a + n_b), and x_b = n_b/(n_a + n_b) = 1 - x_a
Here, we have that n_a = n_b = 2, so x_a = x_b = 1/2, and you are given the molar entropies of the two components, so:
S = 186.19 J/K + 229.49 J/K - (2mol)*(8.314 J/(k*mol))*[0.5*ln(0.5) + 0.5*ln(0.5)]
S = 415.68 J/k - (16.628 J/K)*ln(1/2)
S = 415.68 J/k + (16.628 J/K)*ln(2)
S = 427.206 J/K