Ch. 6 Gases

Gas Laws

$\textcolor{#EE4266}{T}$ : temperature (Kelvin, $\text{Kelvin} = \text{Celcius} + 273.15$ )
$\textcolor{#EE4266}{P}$ : pressure
$\textcolor{#EE4266}{V}$ : volume
$\textcolor{#EE4266}{n}$ : moles
$\textcolor{#EE4266}{R}$ : gas constant
$R=\pu{8.314 J//mol K}=\pu{0.08206 L atm//mol K}=\pu{62.36 L torr// mol K}$

Boyle's Law (1660)

$P_0 V_0=P_1 V_1$

Charle's Law (1787)

$\frac{V_0}{T_0}=\frac{V_1}{T_1}$

Gay-Lussac's Law (c. 1787)

$\frac{P_0}{T_0}=\frac{P_1}{T_1}$

Avogadro's Principle

$\frac{n_0}{V_0}=\frac{n_1}{V_1}$

Ideal Gas Law

$PV=nRT$

STP (or Standard Temperature and Pressure) means $P=\pu{1.00 atm}$ and $T=\pu{273 K}$

Example

A $\pu{12.0 L}$ sample of helium gas at STP is heated to $\pu{100\degree C}$ and pressurized to $\pu{3.00 atm}$ . What is the new volume of helium?

We know $V_0=\pu{12.0 L}$ , $P_0=\pu{1.00 atm}$ , $P_1=\pu{3.00 atm}$ , $T_0=\pu{273 K}$ , $T_1=\pu{373 K}$ , and $n$ and $R$ remain unchanged.
So, from the ideal gas law,
$\frac{P_0 V_0}{T_0}=nR=\frac{P_1 V_1}{T_1}$
Plugging in,
$\frac{1.00\cdot12.0}{273}=\frac{3.00\cdot V_1}{373}$
$V_1=5.47$ , so the new volume is $\pu{5.47 L}$

Density

From $n=\frac{\text{grams of gas}}{\text{molar mass}}$ (look at the units), the density can be found from the ideal gas law (remember $\text{density}=\frac{g}{V}$ ):
$PV=\frac{g}{\text{molar mass}}RT\rightarrow P(\text{molar mass})=\frac{g}{V}RT$

Kinetic Molar Theory

gases consist of atoms/molecules in continuous, random motion
collisions between these atoms/molecules are elastic
the volume occupied is negligibly small
attractive/repulsive forces are negligible
average kinetic energy is directly proportional to the Kelvin temperature of the gas

The highlighted two define an ideal gas

pressure is the amount of force gas particles exert upon colliding with walls per unit area
increasing temperature increases velocity, which increases force $\rightarrow$ higher pressure
decreasing volume increases the frequency that collisions occur $\rightarrow$ higher pressure
Graham's law of effusion:
- $\sqrt{\frac{m_1}{m_2}}=\frac{v_2}{v_1}$
- the rate of effusion (gas going through a hole) is inversely related to the square root of the mass of gas particles

Example

Helium leaks through a hole at a rate of $\pu{3.22*10^-5 mol/s}$ .
How fast will oxygen effuse under the same conditions?

Helium has mass $m_1=4$ and effusion rate $v_1=\pu{3.22*10^-5 mol/s}$
Oxygen has mass $m_2=32$ .
Plugging this in, we have
$\sqrt{\frac{4}{32}}=\frac{v_2}{\pu{3.22*10^-5 mol/s}}\rightarrow v_2=\pu{1.14*10^-5 mol/s}$
So oxygen will effuse at a rate of $\pu{1.14*10^-5 mol/s}$

Real Gas

Volume and attractive/repulsive forces stop being negliglbe under high pressures or low temperatures, so they deviate from the ideal gas law

volume will decrease proportional to the number of moles, so $\text{Volume}=V-nb$ , where $b$ is some proportionality constant
attractions change collision rate, so it was concluded $\text{Pressure}=P+a(\frac{n}{V})^2$
Thus, we have the van der Walls equation:
$(P+\frac{n^2a}{V^2})(V-bn)=nRT$

Dalton's Law of Partial Pressures

When gases are mixed, they tend to act independently
So, total pressure is a combiniation of all gas pressures
$P_{\text{total}}=P_1+P_2+\cdots$
Each $P_k$ is the partial pressure of a gas, and is proportional to number of moles by the ideal gas law