What pressure is exerted by 727.2 g of CH4 in a 0.950 L steel container at 221.9 K ?

Answer 1

Ideal gas law: PV = nRT

Write down what you know and what you want to fine:

P = pressure = ?

V = volume in liters = 0.950 L

n = moles = 727.2 g x 1 mol CH4 / 16 g = 45.45 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 221.9K

Solve for P:

P = nRT/V = (45.45)(0.0821)(221.9) / 0.950

P = 872 atm (3 sig. figs. based on 0.950 L having 3 s.f.)