A closed vessel with a capacity of 2 m3 contains 280 g of nitrogen (= 0.028 kg / mol) and 320 g of oxygen

A closed vessel with a capacity of 2 m3 contains 280 g of nitrogen (= 0.028 kg / mol) and 320 g of oxygen (= 0.032 kg / mol). At a temperature of 16 ° C, the pressure of such a gas mixture in the vessel is equal to …

280 g = 0.28 kg;

320 g = 0.32 kg;

16 ° C = 289 K;

Let’s find the amount of substance gases:

n = m / M;

n (O2) = 0.32 / 0.032 = 10 mol;

n (N2) = 0.28 / 0.028 = 10 mol;

ntotal = n (O2) + n (N2) = 10 + 10 = 20 mol;

Let us find the gas pressure by expressing it from the Mendeleev-Clapeyron equation:

P = (n * R * T) / V;

P = (20 * 8.314 * 289) / 2 = 24027 Pa = 24.027 kPa.

Answer: P = 24.027 kPa.