Before developing the molecular theory of gases, it is useful to review the empirical gas laws from general chemistry. These laws describe how pressure, volume, temperature, and amount of gas are related when one or more variables are held constant. In all gas-law calculations, temperature must be expressed on the Kelvin scale.
Boyle's Law describes the relationship between pressure and volume for a fixed amount of gas at constant temperature. If the temperature and number of moles are held constant, pressure and volume are inversely proportional:
\[ P \propto \frac{1}{V} \qquad \text{or} \qquad P_1V_1 = P_2V_2 \]
Charles' Law describes the relationship between volume and temperature for a fixed amount of gas at constant pressure. If pressure and number of moles are held constant, volume is directly proportional to absolute temperature:
\[ V \propto T \qquad \text{or} \qquad \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Gay-Lussac's Law describes the relationship between pressure and temperature for a fixed amount of gas at constant volume. If volume and number of moles are held constant, pressure is directly proportional to absolute temperature:
\[ P \propto T \qquad \text{or} \qquad \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Avogadro's Law describes the relationship between volume and amount of gas at constant temperature and pressure. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules:
\[ V \propto n \qquad \text{or} \qquad \frac{V_1}{n_1} = \frac{V_2}{n_2} \]
Boyle's, Charles', and Gay-Lussac's Laws can be combined into a single expression for a fixed amount of gas. When the number of moles does not change, the quantity \(PV/T\) remains constant:
\[ \frac{PV}{T} = \text{constant} \]
\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]
When the amount of gas is allowed to change, the empirical gas laws combine to produce the Ideal Gas Law:
\[ PV = nRT \]
In this equation, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(T\) is temperature in Kelvin, and \(R\) is the ideal gas constant. The numerical value of \(R\) depends on the units used for pressure, volume, and energy.
| Gas constant | Units | Use when... |
|---|---|---|
| \(0.08206\) | \(\mathrm{L\,atm\,mol^{-1}\,K^{-1}}\) | Pressure is in atm and volume is in L |
| \(8.314\) | \(\mathrm{J\,mol^{-1}\,K^{-1}}\) | Energy units are needed, or pressure-volume work is expressed in joules |
Big picture: The empirical gas laws are special cases of the Ideal Gas Law. Each law describes how two variables change while the others are held constant.
A sample of gas occupies \(20.0\ \mathrm{L}\) at a pressure of \(1.00\ \mathrm{atm}\). What volume will the gas occupy if the pressure is changed to \(400\ \mathrm{Torr}\), assuming the temperature remains constant?
Because the temperature and amount of gas remain constant, Boyle's Law applies:
\[ P_1V_1 = P_2V_2 \]
Before substituting values, the pressure units must be made consistent. Since the final pressure is given in Torr, convert the initial pressure:
\[ 1.00\ \mathrm{atm} \left( \frac{760\ \mathrm{Torr}} {1\ \mathrm{atm}} \right) = 760\ \mathrm{Torr} \]
The known quantities are therefore
\[ P_1 = 760\ \mathrm{Torr} \]
\[ V_1 = 20.0\ \mathrm{L} \]
\[ P_2 = 400\ \mathrm{Torr} \]
Solving Boyle's Law for the final volume gives
\[ V_2 = \frac{P_1V_1}{P_2} \]
Substituting the known values:
\[ V_2 = \frac{ \left(760\ \mathrm{Torr}\right) \left(20.0\ \mathrm{L}\right) } { 400\ \mathrm{Torr} } \]
The pressure units cancel:
\[ V_2 = \frac{ \left(760\right) \left(20.0\right) } { 400 } \mathrm{L} \]
\[ V_2 = 38.0\ \mathrm{L} \]
Therefore, the gas will occupy
\[ V_2 = 38.0\ \mathrm{L} \]
Physical interpretation: The pressure decreases from \(760\ \mathrm{Torr}\) to \(400\ \mathrm{Torr}\), so the gas expands. Because pressure and volume are inversely proportional, lowering the pressure causes the volume to increase.
Consider the reaction
\[ 3H_2(g) + N_2(g) \rightarrow 2NH_3(g) \]
If \(6.0\ \mathrm{L}\) of \(H_2\) gas is mixed with \(2.0\ \mathrm{L}\) of \(N_2\) gas at the same temperature and pressure, and the reaction goes to completion, how many liters of \(NH_3\) are produced?
Avogadro's Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. As a result, the coefficients in a balanced chemical equation can be interpreted as volume ratios as well as mole ratios.
The balanced equation indicates that
\[ 3\ \mathrm{L}\ H_2 + 1\ \mathrm{L}\ N_2 \rightarrow 2\ \mathrm{L}\ NH_3 \]
Compare the volumes given in the problem to the required stoichiometric ratio:
\[ \frac{6.0\ \mathrm{L}\ H_2} {2.0\ \mathrm{L}\ N_2} = 3.0 \]
This exactly matches the stoichiometric ratio
\[ \frac{3\ \mathrm{L}\ H_2} {1\ \mathrm{L}\ N_2} = 3.0 \]
Therefore, neither reactant is in excess and both are consumed completely.
Use the volume ratio from the balanced equation:
\[ 3\ \mathrm{L}\ H_2 \rightarrow 2\ \mathrm{L}\ NH_3 \]
\[ 6.0\ \mathrm{L}\ H_2 \left( \frac {2\ \mathrm{L}\ NH_3} {3\ \mathrm{L}\ H_2} \right) = 4.0\ \mathrm{L}\ NH_3 \]
Therefore, the reaction produces
\[ 4.0\ \mathrm{L}\ NH_3 \]
Physical interpretation: Because all gases are measured at the same temperature and pressure, the volume ratios are identical to the mole ratios in the balanced chemical equation. This allows gas stoichiometry problems to be solved directly using volumes without first converting to moles.
A sample of gas occupies \(30.0\ \mathrm{L}\) at \(2.00\ \mathrm{atm}\) and \(300\ \mathrm{K}\). What volume will the gas occupy at \(3.00\ \mathrm{atm}\) and \(450\ \mathrm{K}\)?
Predict what will happen to the volume.
Increasing the pressure tends to decrease the volume, while increasing the temperature tends to increase the volume. Which effect do you think will dominate?
How many mL of \(H_2S(g)\) will be produced at \(19.0^\circ\mathrm{C}\) and \(752\ \mathrm{Torr}\) when \(15.0\ \mathrm{mL}\) of \(1.00\ \mathrm{M}\ HCl\) reacts with excess \(FeS(s)\)?
\[ 2HCl(aq) + FeS(s) \rightarrow H_2S(g) + FeCl_2(aq) \]
First calculate the moles of \(HCl\):
\[ n_{HCl} = (0.0150\ \mathrm{L}) \left( 1.00\ \frac{\mathrm{mol}}{\mathrm{L}} \right) = 0.0150\ \mathrm{mol} \]
Use the balanced equation to find the moles of \(H_2S\):
\[ 0.0150\ \mathrm{mol}\ HCl \left( \frac{1\ \mathrm{mol}\ H_2S} {2\ \mathrm{mol}\ HCl} \right) = 0.00750\ \mathrm{mol}\ H_2S \]
Now use the Ideal Gas Law:
\[ PV=nRT \]
\[ V=\frac{nRT}{P} \]
Convert the temperature and pressure:
\[ T = 19.0 + 273.15 = 292.15\ \mathrm{K} \]
\[ P = 752\ \mathrm{Torr} \left( \frac{1\ \mathrm{atm}} {760\ \mathrm{Torr}} \right) = 0.989\ \mathrm{atm} \]
Substitute into the Ideal Gas Law:
\[ V = \frac{ (0.00750\ \mathrm{mol}) \left(0.08206\ \frac{\mathrm{L\,atm}} {\mathrm{mol\,K}}\right) (292.15\ \mathrm{K}) } { 0.989\ \mathrm{atm} } \]
\[ V = 0.182\ \mathrm{L} \]
Convert liters to milliliters:
\[ 0.182\ \mathrm{L} \left( \frac{1000\ \mathrm{mL}} {1\ \mathrm{L}} \right) = 182\ \mathrm{mL} \]
Therefore, the reaction produces
\[ 182\ \mathrm{mL}\ H_2S(g) \]
Key idea: This problem combines solution stoichiometry with the Ideal Gas Law. The \(HCl\) determines the moles of gas produced, and the Ideal Gas Law converts those moles into a gas volume at the stated temperature and pressure.
Big picture: The empirical gas laws describe how pressure, volume, temperature, and amount of gas are related. The Ideal Gas Law unifies these relationships into a single equation that serves as the foundation for the theoretical treatment of gases developed in the remainder of the chapter.