Question

How many moles of nitrogen gas will occupy a volume of 347 mL at 6700 kPa and 27 degrees Celsius

Answer 1

1) List the known and unknown quantities.

Volume: 347 mL.

Pressure: 6700 kPa.

Temperature: 27 ºC.

Ideal gas constant: 8.314 L * kPa * K^(-1) * mol^(-1).

Moles: unknown.

2) Set the equation.

Ideal gas equation

$PV=nRT$

3) Convert units.

3.1- Convert the volume.

1L = 1000 mL

$L=347\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.347\text{ }L$

3.2- Convert the temperature

$K=ºC+273.15$$K=27+273.15=300.15\text{ }K$

4) Plug in the known quantities (ideal gas equation) and solve for n (moles).

$(6700\text{ }kPa)(0.347\text{ }L)=n*(8.314\text{ }L*kPa*K^{-1}*mol^{-1})(300.15\text{ }K)$$n=\frac{(6700\text{ }kPa)(0.347L)}{(8.314\text{ }L*kPa*K^{-1}*mol^{-1})(300.15\text{ }K)}$$n=0.9317\text{ }mol$

There would be 0.9317 mol N2.