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