Question

A nitrogen gas container has a volume of 35.0 L. Calculate the mass of nitrogen in the container if the gas has a pressure of 1800 torr at 42°C

Answer 1

Q265203

A nitrogen gas container has a volume of 35.0 L. Calculate the mass of nitrogen in the container if the gas has a pressure of 1800 torr at 42°C

Solution :

Step 1: To find the moles of N₂ gas in the container.

The volume of nitrogen gas in the container, V = 35.0 L.

Pressure, P = 1800 torr.

convert pressure to 'atm' by using the conversion factor, 1atm = 760 torr.

Pressure, P in 'atm' =

$$= 1800 \ \cancel{torr} * \frac{1 \ atm }{760 \ \cancel{torr} }$$

$$=2.368 \ atm$$

Temperature, T in 'Kelvin' = 42°C = 42 + 273.15 = 315.15 K.

Gas constant, R = 0.08206 L-atm/mol-K

plug all this information in the ideal gas equation, PV = n RT, and solve it for 'n'.

$$P \ V = \ n \ R \ T$$

$$2.368 \ atm \ * 35.0 \ L \ = \ n \ * 0.08206 L.atm/mol.K * 315.15 \ K$$

$$82.88 \ L.atm = \ n \ * 25.861 \ L.atm/mol.$$

divide both the side by 25.861 L.atm/mol, we have

$$\frac{82.88 \ \cancel{L.atm}}{25.861 \ \cancel{L.atm}/mol} = \frac{\ n \ * \cancel{25.861 \ L.atm/mol}}{\cancel{25.861 \ L.atm/mol}}.$$

3.205 \ mol = \ n \

Step 2 : To convert 3.205 mol N₂ to grams.

Molar mass of N₂ = 2 * atomic mass of N = 2 * 14.007 g/mol

= 28.014 g/mol

$$grams \ of \ N_2 = 3.205 \ \cancel{mol \ N_2} * \frac{28.014 \ g \ N_2 }{ 1 \ \cancel{mol \ N_2}} = 89.78 \ grams \ of \ N_2$$

which is the correct significant figure is 89.8 grams.

Hence the mass of N₂ in the container is 89.8 grams.