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

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

Answer 2

The quantity of nitrogen the cylinder contains is 4477.8 g

What is pressure?

Pressure is a force exerted in a perpendicular direction in any item.

By ideal gas law

PV = nRT

$P =\dfrac{w}{M} =\dfrac{RT}{V}$

w = mass

Volume is 20.0 l

Pressure is  $2.00 \times 10^4 \;Kpa$

The molar mass of nitrogen is 28 g/mol

R is gas constant = 0.0821

Temperature is 28 converted into kelvin that is 301 k

Putting the values

$197.6 =\dfrac{w}{28} =\dfrac{0.0821 \times 301 }{20\;l}\\\\w = 4477.8 g$

Thus, the mass of nitrogen is 4477.8 g.

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