Liquid and vapor phases of benzene are in equilibrium at 353.2 K. Calculate the molar volume of benzene in vapor pressure. Use the following data in your calculation.Avap H = 30.8 kJ mol¹; dP/dT = 0.0437 atm K-¹Molar volume of benzene at 353.2 K = 96.0 cm³ mol-¹Answer Choices:(A) 29.0 L mol-¹(B) 29.70 L mol-1(C) 19.79 L mol-¹(D) 96.79 L mol-1

To solve this problem, we need to use the Clausius-Clapeyron equation, which is:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Given that the liquid and vapor phases of benzene are in equilibrium, we can assume that P2 = P1. Therefore, the equation simplifies to:

ln(1) = -ΔHvap/R * (1/T2 - 1/T1)

Since ln(1) = 0, we can further simplify the equation to:

0 = -ΔHvap/R * (1/T2 - 1/T1)

Rearranging the equation gives us:

ΔHvap/R = (1/T2 - 1/T1)

We know that ΔHvap = 30.8 kJ/mol = 30800 J/mol, R = 8.314 J/(mol*K), and T1 = T2 = 353.2 K. Substituting these values into the equation gives us:

30800 J/mol / 8.314 J/(mol*K) = (1/353.2 K - 1/353.2 K)

Solving the equation gives us:

3703.6 K = 0

This is a contradiction, which means that our initial assumption that P2 = P1 is incorrect. Therefore, we cannot use the Clausius-Clapeyron equation to solve this problem.

Instead, we need to use the ideal gas law, which is:

PV = nRT

We know that P = 1 atm, V is the molar volume we want to find, n = 1 mol, R = 0.0821 Latm/(molK), and T = 353.2 K. Substituting these values into the equation gives us:

1 atm * V = 1 mol * 0.0821 Latm/(molK) * 353.2 K

Solving the equation gives us:

V = 0.0821 Latm/(molK) * 353.2 K

V = 28.98 L/mol

Therefore, the molar volume of benzene in vapor phase at 353.2 K is approximately 29.0 L/mol. The correct answer is (A) 29.0 L mol-¹.