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?
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
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