Establish the pressure of 500 g of ethylene with a volume of 20 liters at a temperature of -2 ° C.

Given:
m (C2H4) = 500 g
V1 (C2H4) = 20 l
T1 = -2оС
Pн = 101.3 kPa
Tn = 273K

To find:
P1 (C2H4) -?

Decision:
1) M (C2H4) = Mr (C2H4) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 2 + 1 * 4 = 28 g / mol;
2) n (C2H4) = m (C2H4) / M (C2H4) = 500/28 = 17.86 mol;
3) Vn (C2H4) = n (C2H4) * Vm = 17.86 * 22.4 = 400.06 L;
4) T1 = 273 – 2 = 271K;
5) Pn * Vn / Tn = P1 * V1 / T1;
6) P1 = (Pn * Vn * T1) / (Tn * V1) = (101.3 * 400.06 * 271) / (273 * 20) = 2011.5 kPa.

Answer: The pressure of C2H4 is 2011.5 kPa.