At a temperature of -20 s and a pressure of 1.33 * 10 ^ 5 Pa, the gas occupies a volume of 2 liters.

At a temperature of -20 s and a pressure of 1.33 * 10 ^ 5 Pa, the gas occupies a volume of 2 liters. Find the volume of gas under normal conditions.

Data: Т1 (considered temperature) = -20 ºС (253 К); P1 (considered pressure) = 1.33 * 10 ^ 5 Pa; V1 (considered volume) = 2 liters.

Constants: P2 (normal atmospheric pressure) = 101325 Pa; T2 (standard temperature) = 273 K.

To find the required gas volume under normal conditions, we use the proportion: P1 * V1 / T1 = P2 * V2 / T2 and V2 = P1 * V1 * T2 / (T1 * P2).

Calculation: V2 = 1.33 * 10 ^ 5 * 2 * 273 / (253 * 101325) = 2.83 liters.

Answer: Under n.u. gas will occupy a volume of 2.83 liters.