Answer : The mass of \(N_2\) gas is, 4477.8 g
Solution :
using ideal gas equation,
\(PV=nRT\\\\P=\frac{w}{M}\times \frac{RT}{V}\)
where,
n = number of moles of gas
w = mass of gas
P = pressure of the gas = \(2\times 10^4Kpa=197.6atm\)
conversion : \(1atm=101.2Kpa\)
T = temperature of the gas = \(28^oC=273+28=301K\)
M = molar mass of \(N_2\) gas = 28 g/mole
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = 20 L
Now put all the given values in the above equation, we get the mass of gas.
\(197.6atm=\frac{w}{28g/mole}\times \frac{0.0821Latm/moleK\times 301K}{20L}\)
\(w=4477.8g\)
Therefore, the mass of \(N_2\) gas is, 4477.8 g
