What is the total pressure in a 6.00-l flask which contains 0.127 mol of h2(g) and 0.288 mol of n2(g) at 20.0°c?

1.66 atm

Data:

P = ?
V = 6.00 liter
n1 = 0.127 mol of H2(g)
n2 = 0.288 mol of N2(g)
T = 20.0°C

Formula:

PV = n RT

Solution:

1) Conversion of units:

 =  20 + 273.15 K = 293.15 K

2) n = n1 + n2 = 0.127 mol + 0.288 mol = 0.415 mol

3) PV = nRT => P = nRT / V

R = 0.0821 atm*liter / (K*mol)

P = 0.415 mol * 0.0821 atm * liter / (K*mol) * 293.15 K / 6 liter = 1.66 atm

1.66 atm

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