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?

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