Question

What pressure is exerted by 920.6 g of CH4 in a 0.550 L steel container at 129.6 K?

Answer 1

Given mass of CH₄ , m = 920.6 g

Molar mass of CH₄ = At.mass of C + (4xAt.mass of H)

                            = 12 + (4x1)

                            = 16 g/mol

So number of moles , n = mass /molar mass

                                  = 920.6 g / 16(g/mol)

                                    = 57.54 mol

We know that PV = nRT

Where

T = Temperature = 129.6 K

P = pressure = ?

n = No . of moles = 57.54 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 0.550L

Plug the values we get

P = (nRT) / V

   = ( 57.54 x0.0821x129.6) / 0.550

   = 1113 atm

Therefore the pressure exerted is 1113 atm