What volume is needed to store 0.80 moles of helium gas at 204.6 kPa and 300 K?

Jun 10, 2016

The volume of helium gas is $9.76 L$

Explanation:

For this type of question we would use the ideal gas law equation
$P \times V = n \times R \times T$.

• P represents pressure (must have units of atm)
• V represents volume (must have units of liters)
• n represents the number of moles
• R is the proportionality constant (has a value of 0.0821 with units of $\frac{L \times a t m}{m o l \times K}$)
• T represents the temperature, which must be in Kelvins.

Now what you want to do is list your known and unknown variables. Our only unknown is the volume of helium gas. Our known variables are P,n,R, and T.

The pressure has the incorrect units because the pressure should be in atmospheres instead of kPa. In order to go from kPa to atm we use the following relationship:

$101.325 k P a = 1 a t m$

$\left(204.6 \cancel{\text{kPa")/(101.325cancel"kPa}}\right) \times 1 a t m$ = $2.019 a t m$

Now all we have to do is rearrange the equation and solve for P like so:
$V = \frac{n \times R \times T}{P}$
$V = \left(0.80 \cancel{\text{mol"xx0.0821(Lxxcancel"atm")/(cancel"mo"lxxcancel"K")xx(300cancel"K"))/(2.019cancel"atm}}\right)$
$V = 9.76 L$