Calculate the entropy change when 2.0 mol of a perfect gas A and 3.0 mol of a pe

Question

Calculate the entropy change when 2.0 mol of a perfect gas A and 3.0 mol of a perfect gas B mix spontaneously. -12 JK^-1 -8.3 kJ k^-1 +195 JK^-1 +28 JK^-1 The boiling temperature of ethylbenzene is 136 degree C. Use Trouton's rule to estimate the enthalpy of vaporization of ethylbenzene at this temperature. 2.3 kJ mol^-1 4.8 kJ mol^-1 35 kJ mol^-1 12 kJ mol^-1 At 90 degree C, the vapour pressure of 1, 2-dimethylbenzene is 20 kPa and that of 1, 3-dimethylbenzene is 18 kPa. What is the mole fraction of 1, 2-dimethylbenzene in vapour of a liquid mixture that boils at 90 degree C when the pressure is 19 kPa? 0.900 0.500 0.526 0.474

Explanation / Answer

Answer (13)

When perfect gases A and B mix, the entropy change is given by eqn delta (S) = -nR (xAlnxA +(xBlnxB ).

If the total number of moles is

n = nA + nB =2.0 mol + 3.0 mol = 5.0 mol

then the mole fraction of A and B are

xA = nA / n = 2.0 mol / 5.0 mol = 0.4

xB = nB / n = 3.0 mol / 5.0 mol = 0.6

thus

delta (S) = -5.0 mol x 8.1345 J K-1 x x { (0.4 ln 0.4) + (0.6 ln 0.6)}

delta (S) = 28 J / K = (d) Answer

Answer (14)

Trounton's rule is valid for many hydrocarbon and states that

delta(H0) (Tb) / Tb = 85 J /K*mol

Thus

delta(H0) (Tb) = 85 x (273.15 + 13)

delta(H0) (Tb) = 35 x 10^2 J / mol

delta(H0) (Tb) = 35 kJ / mol = (c) Answer

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.