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

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

Step-by-step 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* 10^(3)\right)\; \rm Pa`.
• 
  `V = 520\; \rm mL = \left(520 * 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 &= (P \cdot V)/(R \cdot T) \\&= (\left(150 * 10^3\right)\; \rm Pa * \left(520 * 10^(-6)\right)\; \rm m^(-3))/(8.314\; \rm m^3\cdot Pa \cdot K^(-1)\cdot mol^(-1) * 298.15\; \rm K)\\ &\approx 3.2559 * 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 * 15.999 = 44.009\; \rm g \cdot mol^(-1)`.

The mass of that
`3.2559 * 10^(-2)\; \rm mol` of
`\rm CO_2` in this balloon would be:

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