Question

A balloon is at 15oC,
150 kPa, and is 520
mL in volume.
Calculate how
many grams of CO2
gas are in the balloon

Answer 1

Answer:

Approximately $1.4\; \rm g$, assuming that $\rm CO_2$ acts like an ideal gas.

Explanation:

Number of moles of CO₂ molecules in the balloon

Look up the ideal gas constant:

$R = 8.314\; \rm m^3\cdot Pa \cdot K^{-1}\cdot mol^{-1}$.

Convert the unit of temperature, pressure, and volume to standard units:

• $T = 15\; \rm ^\circ C = (15 + 273.15) \; \rm K = 298.15\; \rm K$.
• $P = 150\; \rm kPa = \left(150\times 10^{3}\right)\; \rm Pa$.
• $V = 520\; \rm mL = \left(520 \times 10^{-6}\right)\; \rm m^{-3}$

Assume that the $\rm CO_2$ here acts like an ideal gas. Apply the ideal gas law to find the number of moles of

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\&= \frac{\left(150 \times 10^3\right)\; \rm Pa \times \left(520 \times 10^{-6}\right)\; \rm m^{-3}}{8.314\; \rm m^3\cdot Pa \cdot K^{-1}\cdot mol^{-1} \times 298.15\; \rm K}\\ &\approx 3.2559 \times 10^{-2}\; \rm mol \end{aligned}$.

Mass of the CO₂ in the balloon

Look up the following relative atomic mass data on a modern periodic table:

• $\rm C$: $\rm 12.011$.
• $\rm O$: $\rm 15.999$.

Calculate the molar mass of $\rm CO_2$:

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$.

The mass of that $3.2559 \times 10^{-2}\; \rm mol$ of $\rm CO_2$ in this balloon would be:

$\begin{aligned}m &= n \cdot M \\ &= 3.2559 \times 10^{-2}\; \rm mol \times 44.009 \; \rm g \cdot mol^{-1} \\ &\approx 1.4\; \rm g\end{aligned}$.