# Density of ammonia gas in a 4.32L container at 45 C and 837 torr…is ?

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• PV = nRT

n = mass / mw

PV = (mass / mw) RT

(mass / V) = mw x P / (RT)

density = mw x P / (RT)

density = (17.0 g/mole) x (837/760 atm) / [(0.0821 Latm/moleK) x (328K)]

density = 0.717 g/L

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just for the record…ammonia’s not really the best example to use for this type of problem. Because it has a high degree of hydrogen bonding and tends to deviate from ideality.

If I were to do the same type of analysis via Van Der Waals equation, which includes a term for intermolecular forces, I’d find this…

(P + an² / V²) x (V – nb) = nRT

where

P = 837 /760 = 1.101 atm

a = 4.170 L² atm / mole²

b = 0.03707 L/mole

R = 0.08206 Latm/moleK

T = 273.15 + 45 = 328 K

and if I let V = 1 L…

(1.101 atm + 4.170 L² atm / mole² n² / (1L)²) x (1L – 0.03707 L/mole n) = 0.08206 Latm/moleK x 328 K x n

factoring out atm, L, canceling L², simpilfying the right side…

(1.101 + 4.170 / mole² n² / ) x (1 – 0.03707 /mole x n) = 26.92 /mole x n

so the units will end up wth n in moles.. so I’ll cancel the units for simplicity…

(1.101 + 4.170 n² ) x (1 – 0.03707 n) = 26.92 x n

1.101 – 0.0408 n + 4.170 n² – 0.1546 n³ = 26.92 n

0.1546 n³ – 4.170 n² + 26.96 n – 1.101 = 0

the solution to that cubic is n = 0.04086

so that mass NH3 = 0.04086 moles x (17.0 g / mole) = 0.695 g per the 1 L that I assumed…

so density = 0.695 g/L… difference is about 3%…

• Convert pressure to atm, and T to Kelvin:

P = 837/760 = 1.10 atm

T= 45+273 = 318 K

PV = nRT

Rearrange ideal gas law to give:

n/V = P/RT = 1.10 atm/(.0821 Latm/molK) 318K = 0.0421 mol/L

multiply mol/L X g/mol to get g/L

0.0421 mol/L X 17 g/mol = 0.716 g/L

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