Octane (C8H18, 114.23 g/mol). The density of octane vapour 125.0 Celsius and 720.0 torr is ? and what would be the volume be if it was 2.00 moles ?

Question

Octane (C8H18, 114.23 g/mol). The density of octane vapour 125.0 Celsius and 720.0 torr is ? and what would be the volume be if it was 2.00 moles ?

1 Answer

Impact of this question

8112 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-08T19:32:24

To determine the density of octane vapor at that temperature and pressure you basically have to use the ideal gas law equation, $P V = n R T$ $PV = nRT$ .

Notice that this form, $P V = n R T$ $PV = nRT$ , is not very useful in your case because you don't have a volume and a number of moles of gas; this means that you must rearrange the equation to incorporate what you have, molar mass, pressure, and temperature.

So, the number of moles of a substance is defined as the ratio between mass and molar mass, which means that

$P V = n R T \Rightarrow P V = \frac{m}{M_{M}} \cdot R T$ $PV = nRT =>PV = m/M_M * RT$

If you rearrange this to isolate the unknown mass and volume on one side, you'll get

$P V \cdot M_{M} = m \cdot R T \Rightarrow P \cdot M_{M} = \frac{m}{V} \cdot R T$ $PV * M_M = m * RT => P * M_M = m/V * RT$

Since density is defined as mass per unit of volume, the above equation will become

$P \cdot M_{M} = ρ \cdot R T \Rightarrow ρ = \frac{P \cdot M_{M}}{R T}$ $P * M_M = rho * RT => rho = (P * M_M)/(RT)$

Plug all your values into this equation (do not forget to use atm and Kelvin)

$ρ = \frac{\frac{720.0}{760} atm \cdot 114.23 \frac{g}{mol}}{0.082 \frac{L \cdot atm}{mol \cdot K} \cdot (273.15 + 125) K} = 3.315 g/L$ $rho = (720.0/760"atm" * 114.23"g"/"mol")/(0.082("L"* "atm")/("mol" * "K") * (273.15 + 125)"K") = "3.315 g/L"$

SIDE NOTE Vapor density is usually given in relation to the vapor density of air at those same values for temperature and pressure. You can determine the vapor density of air at 125 Celsius and 720 torr by using its molar mass, 28.97 g/mol, in the above equation.

You'll get 0.8406 g/L as the vapor density of air, which will make the vapor density of octane - in this context called specific gravity - equal to

$S G_{octane} = \frac{3.315 g/L}{0.8406 g/L} = 3.944$ $SG_("octane") = "3.315 g/L"/"0.8406 g/L" = "3.944"$ $\to$ $->$ UNITLESS

To determine the volume use the classic-form $P V = n R T$ $PV = nRT$ , since now you've got all the data you need

$P V = n R T \Rightarrow V = \frac{n R T}{P} = \frac{2.00 moles \cdot 0.082 \frac{L \cdot atm}{mol \cdot K} \cdot (273.15 + 125) K}{\frac{720.0}{760} atm}$ $PV = nRT => V = (nRT)/P = ("2.00 moles" * 0.082("L"* "atm")/("mol" * "K") * (273.15 + 125)"K")/(720.0/760"atm")$

$V = 68.9 L$ $V = "68.9 L"$

This time, the answer is rounded to three sig figs because of the number of sig figs in 2.00.

Science

Math

Humanities

... and beyond

Octane (C8H18, 114.23 g/mol). The density of octane vapour 125.0 Celsius and 720.0 torr is ? and what would be the volume be if it was 2.00 moles ?

1 Answer

Related questions

Impact of this question