Question

If 50.9 g of NH3 occupies 35.1 L under a pressure of 83.9 in. Hg, what is the temperature of the gas, in °C?

Answer 1

Mass of NH3 = 50.9 g, Molar mass of NH3 = 17.03 g/mol

Number of moles of NH3 = 50.9 g/ 17.03 g/mol = 2.99 mol

Volume = 35.1 L,

Pressure = 83.9 in.Hg = 83.9 * 0.0334 atm = 2.8 atm

1 in Hg = 0.0334 atm

We know from gas law equation , PV = n * R * T -----(1)

Where, P = Pressure in atm, V = Volume in Litre

n = number of moles

R = gas constant = 0.082 atm L mol-1K-1

T = Temperure in Kelvin

From (1)

T = PV / n * R

= 2.8 atm * 35.1 L / 2.99 mol * 0.082 atm L mol-1K-1

= 98.28 K/ 0.2452 = 400.81 K

Temperature in ^oC = (400.81 - 273.15) ^oC = 127.66 ^oC