Question

The gas SF6 is used to trace air flows because it is non-toxic and can be detected selectively in air at a concentration of 1.0 ppb. What partial pressure is this? At this concentration, how many molecules of SF6 are contained in 1.0 cm3 of air at T = 46 °C

Answer 1

SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.

First, we will calculate the partial pressure of SF₆  using the following expression.

$pSF_6 = P \times \frac{ppb}{10^{9} }$

where,

• pSF₆: partial pressure of SF₆

• P: total pressure of air (we will assume it is 1 atm)

• ppb: concentration of SF₆ in parts per billion

$pSF_6 = P \times \frac{ppb}{10^{9} } = 1 atm \times \frac{1.0 ppb}{10^{9} } = 1.0 \times 10^{-9} atm$

Then, we will convert 1.0 cm³ to L using the following conversion factors:

• 1 cm³ = 1 mL
• 1 L = 1000 mL

$1.0 cm^{3} \times \frac{1mL}{1cm^{3}} \times \frac{1L}{1000 mL} = 1.0 \times 10^{-3} L$

Next, we will convert 46 °C to Kelvin using the following expression.

$K = \° C + 273.15 = 46 + 273.15 = 319 K$

Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.

$P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.0 \times 10^{-9} atm \times 1.0 \times 10^{-3} L}{(0.082 atm.L/mol.K) \times 319 K} = 3.8 \times 10^{-14} mol$

Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.

$3.8 \times 10^{-14} mol \times \frac{6.02 \times 10^{23} molecules }{mol} = 2.3 \times 10^{10} molecules$

SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.

You can learn more about partial pressure here: brainly.com/question/13199169