Question

A 25 L cylinder contains 128 g of nitrogen gas at 10oC. How many grams of nitrogen must be added to increase the pressure to 5.00 atm assuming ideal gas behavior

Answer 1

Answer : The amount of nitrogen added must be, 22.64 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 5.00 atm

V = Volume of $N_2$ gas = 25 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $10^oC=273+10=283K$

Putting values in above equation, we get:

$5.00atm\times 25L=n\times (0.0821L.atm/mol.K)\times 283K$

$n=5.38mol$

Now we have to calculate the mass nitrogen gas.

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

Molar mass of nitrogen gas = 28 g/mol

$\text{Mass of }N_2=5.38mol\times 28g/mol=150.64g$

Now we have to calculate the amount of nitrogen must be added.

Amount of nitrogen must be added = 150.64 g - 128 g

Amount of nitrogen must be added = 22.64 g

Thus, the amount of nitrogen added must be, 22.64 grams.