Question

CS2 is a commonly used nonpolar solvent and has a vapor pressure of 0.45 MPa at 100C and 0.782 MPa at 125C.

A. Calculate the normal boiling temp of CS2 based on the given pressure data.

B. Use the given vapor pressures to estimate rhea centric factor for CS2 whose critical temp and pressure are 552 K and 7.80 MPa.

Answer 1

A).

Let us use the Clausius-Clapeyron Equation:

ln (P₁ / P₂) = (ΔH / R) (1/T₂ - 1/T₁) where ΔH = heat of vaporization in J/mol

Here P1 = 0.45MPa, T1 = 100 + 273 = 373 K

and P2 = 0.782, T2 = 125 +273 = 398 K

and R = 8.31

hence, ln (0.45/ 0.782) = (ΔH / 8.31) (1/398 - 1/373)

or -0.553 = (ΔH / 8.31) x (-1.68 x 10^-4)

therefore ΔH = 27354 J/mol

Now let the normal boiling temp. be T, then Vap. Presure at T is 0.1 MPa i.e atmospheric pressure

Hence

ln (0.45/ 0.1) = (27354 / 8.31) (1/T - 1/373)

or (1/T - 1/373) = 4.57 x 10^-4.

or 1/T = 3.14 x 10^-3.

therefore T = 318.47 K = 45.47 ^oC.

B) The acentric factor is defined as:

where P^c is the critical pressure and P_σ is the vapor pressure at temperature T where T/T^c = 0.7 and T^c is the critical temperature.

Here T = T^c x 0.7 = 552 x 0.7= 386.4 K

hence

ln (0.45/ P) = (27354 / 8.31) (1/386.4 - 1/373)

or ln (0.45/ P) = -0.306

therefore 0.45/ P = e^-0.306 = 0.736

so, P = 0.45/ 0.736 = 0.611 MPa

Therefore acentric factor = -1.00 - log₁₀(0.611/ 7.80) = -1.00 - (-1.11) = 0.11