Question

A 250 l container with 2.86 kg of propane gas (c3h8) is at a temperature of 30.2 ºc. what is the pressure (atm) of the gas?

Answer 1

Data Given:
Pressure = P = ?

Volume = V = 250 L

Temperature = T = 30.2 °C + 273 = 303.2 K

Mass = m = 2.86 Kg = 2860 g

Moles = n = 2860 g / 44.1 g.mol⁻¹ = 64.85 moles

Solution:
Let suppose the gas is acting Ideally, Then According to Ideal Gas Equation.

P V = n R T
Solving for P,
P = n R T / V

Putting Values,

P = (64.85 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 303.2 K) ÷ 250 L

P = 6.45 atm