Question

What pressure is exerted by 777.0 g of CH4 in a 0.870 L steel container at 277.1 K?

Answer 1

Molar mass of CH4 = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass of CH4 = 777.0 g

we have below equation to be used:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(777.0 g)/(16.042 g/mol)

= 48.44 mol

we have:

V = 0.87 L

n = 48.4354 mol

T = 277.1 K

we have below equation to be used:

P * V = n*R*T

P * 0.87 L = 48.4354 mol* 0.0821 atm.L/mol.K * 277.1 K

P = 1267 atm

Answer: 1267 atm