To calculate the density of gaseous sulfur hexafluoride (SF6) at a temperature of 15°C and a pressure of 0.942 atm, we can use the ideal gas law:
PV = nRT
where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹)
T is the temperature (in K)
We can rearrange this equation to solve for the density (ρ):
ρ = n/V = P/(RT) * M
where M is the molar mass of SF6 (146.06 g mol⁻¹).
First, we need to convert the temperature to Kelvin:
T = 15°C + 273.15 = 288.15 K
Now, we can plug in all of the known values to calculate the density of SF6:
ρ = (0.942 atm) / (0.08206 L atm mol⁻¹ K⁻¹ * 288.15 K) * 146.06 g mol⁻¹
ρ = 589.588 g/L
Therefore, the density of gaseous SF6 at a temperature of 15°C and a pressure of 0.942 atm is 589.588 grams per liter.