Question

You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2.00 *10^4 Kpa at 28c How many kilograms of N2 does the cylinder contain

Answer 1

This problem can easily be solved using the ideal gas equation which states:

PV = nRT

where:
P = pressure
V = volume
n = no. of moles
R = gas constant
T = absolute temperature

The problem already gives the pressure, volume, and temperature of the system. All that's missing is the number of moles which is related to the mass using:

mass = moles x molecular mass

The R that should be used must have units that are consistent with the given data. The value used is 8314.34 L-Pa/mol-K. The answer is 4.475 kg of N2.

Answer 2

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