# A carbon dioxide sample weighing 44.0 g occupies 32.68 L at 65°C and 645 torr. What is its volume at STP?

##### 1 Answer
Jun 24, 2018

Its volume at STP is 28.09L.

#### Explanation:

We first need to find the volume of the $C {O}_{2}$ sample at the initial conditions.

To do this, we can use the ideal gas law equation, $P V = n R T$, where $P$ = pressure, $V$ = volume, $n$ = moles of gas, $R$ is the ideal gas constant, and $T$ = temperature in Kelvins.

We should first convert our units to standard gas units.
$P$ = 645 Torr = 86.0 kPa.
$T$ = 65 C = 338K.

We also need the moles of gas, $n$. To do this, we need the molar mass of $C {O}_{2}$, which is 44.01 g/mol.
$n = \frac{m}{M}$
$n = \frac{44.0 g}{44.01 \frac{g}{m o l}}$
$n = 1.00 m o l$

The ideal gas constant, $R$, is equal to $8.314 \frac{L \cdot k P a}{m o l \cdot K}$.

Substitute these values into the ideal gas law equation, $P V = n R T$:

$\left(86.0 k P a\right) V = \left(1.00 m o l\right) \left(8.314 \frac{L \cdot k P a}{m o l \cdot K}\right) \left(338 K\right)$.

Rearrange for V:
$V = \frac{\left(1.00 m o l\right) \left(8.314 \frac{L \cdot k P a}{m o l \cdot K}\right) \left(338 K\right)}{86.0 k P a}$
$V = 32.66 L$.

We can call this the initial volume, ${V}_{i}$.

We can now use any of the gas law equations to solve the problem. Let's use Boyle's Law:
${P}_{i} {V}_{i} = {P}_{f} {V}_{f}$.

We can substitute in ${V}_{i} = 32.66 L$, ${P}_{i} = 86.0 k P a$. At STP (standard temperature and pressure), $P = 100.0 k P a$, so we also substitute ${P}_{f} = 100.0 k P a$.
$\left(86.0 k P a\right) \left(32.66 L\right) = \left(100.0 k P a\right) {V}_{f}$

Rearrange for ${V}_{f}$:
${V}_{f} = \frac{\left(86.0 k P a\right) \left(32.66 L\right)}{100.0 k P a}$
${V}_{f} = 28.09 L$.

The final volume is 28.09L.