Question

A gas in a vessel at 120oF and 13.8 psia consists of 2 % nitrogen, 79 % methane, and 19 % ethane.

a) If the ethane were removed from the gas, what would be the pressure in the vessel?

b) What would be the subsequent partial pressure of nitrogen?

Answer 1

(°F − 32) × 5/9 = 0°C So temp = 48.890°C , pressure = 13.8psia = 0.94atm.

Assuming total weight to be 100gm

So moles of N2 = 2/28 = 0.07

Moles of methane = 79/16 = 4.94

Moles of ethane = 19/30 = 0.63 So total moles = 0.63+4.94+0.07 = 5.64

So applying PV = nRT ; V and T is constant so Pn; P/n = constant

a) n_i = 5.64, P_i = 0.94atm n_f = 5.64-0.63 = 5.01(now ehtane is removed), P_f = ?

P_i/n_i = P_f/n_f ;   P_f = 0.835atm = 12.27psia

b) partial pressure of nitrogen = P_T*(X_N2) = 12.27*(0.07/5.01) = 0.17psia