# What is the volume of a container that holds 25.0 g of carbon gas at STP?

Jul 1, 2016

The container has a volume of $46.6 L$.

#### Explanation:

Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation

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

Next, list your known and unknown variables. Our only unknown is the volume of $C \left(g\right)$. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 $\frac{L \times a t m}{m o l \times K}$

The only issue is the mass of $C \left(g\right)$, we need to convert it into moles of $C \left(g\right)$ in order to use the ideal gas law.

25.0cancel"g" xx (1molC)/(12.01cancel"g") = 2.08 mol $C$

Now all we have to do is rearrange the equation and solve for V like so:
$V = \frac{n \times R \times T}{P}$
$V = \frac{2.08 m o l \times 0.0821 \frac{L \times a t m}{m o l \times K} \times \left(273 K\right)}{1 a t m}$
$V = 46.6 L$