Learning Objectives
- Describe the ideal gas law.
- Use the ideal gas law to calculate pressure, volume, temperature, or moles of an ideal gas.
So far, the gas laws we have considered have all required that the gas change its conditions; then we can predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time? Consider a further extension of the combined gas law to include n. By analogy to Avogadro's law, n is positioned in the denominator of the fraction, opposite the volume. So,
\[\frac{PV}{nT}=constant\nonumber \]
Because pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant. Indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R, so the previous equation is written as
\[\dfrac{PV}{nT}=R\nonumber \]
which is usually rearranged as
\[PV = nRT\nonumber \]
This equation is called the ideal gas law. It relates the four independent properties of a gas at any time. The constant \(R\) is called the ideal gas law constant. Its value depends on the units used to express pressure and volume.
Table \(\PageIndex{1}\): Values of the Ideal Gas Law Constant lists the numerical values of \(R\).
| Numerical Value |
Units |
| 0.08205 |
\(\dfrac{L \cdot atm}{mol \cdot K}\) |
| 62.36 |
\(\dfrac{L \cdot torr}{mol \cdot K}=\dfrac{L \cdot mmHg}{mol \cdot K}\) |
| 8.314 |
\(\dfrac{J}{mol \cdot K}\) |
The ideal gas law is used like any other gas law, with attention paid to the unit and expression of the temperature in kelvin. However, the ideal gas law does not require a change in the conditions of a gas sample. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property.
Example \(\PageIndex{1}\)
A 4.22 mol sample of \(\ce{Ar}\) has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?
Solution
The first step is to convert temperature to kelvins:
\[34 + 273 = 307\, \ce{K} \nonumber\]
Now we can substitute the conditions into the ideal gas law:
\[(1.21atm)(V)=(4.22\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(307\, K)\nonumber \]
The atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume that we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21:
\[V=\dfrac{(4.22)(0.08205)(307)}{1.21}L\nonumber \]
Then solving for volume, we get V = 87.9 L
Exercise \(\PageIndex{1}\)
A 0.0997 mol sample of \(\ce{O2}\) has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?
- Answer
-
3.94 L
Example \(\PageIndex{2}\)
At a given temperature, 0.00332 g of Hg in the gas phase has a pressure of 0.00120 mmHg and a volume of 435 L. What is its temperature?
Solution
We are not given the number of moles of Hg directly, but we are given a mass. We can use the molar mass of Hg to convert to the number of moles.
\[0.00332\cancel{g\, Hg}\times \frac{1\, mol\, Hg}{200.59\cancel{g\, \, Hg}}=0.0000165\, mol=1.65\times 10^{-5}mol\nonumber \]
Pressure is given in units of millimeters of mercury. We can either convert this to atmospheres or use the value of the ideal gas constant that includes the mmHg unit. We will take the second option. Substituting into the ideal gas law,
\[(0.00332\, mm\, Hg)(435\, L)=(1.65\times 10^{-5}mol)(62.36\frac{L.mmHg}{mol.K})T\nonumber \]
The mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature. Isolating T on one side, we get
\[T=\frac{(0.00332)(435)}{(1.65\times 10^{-5})(62.36)}K\nonumber \]
Then solving for K, we get T = 1,404 K.
Exercise \(\PageIndex{2}\)
For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?
- Answer
-
8.21 L
The ideal gas law can also be used in stoichiometry problems.
Example \(\PageIndex{3}\)
What volume of \(\ce{H2}\) is produced at 299 K and 1.07 atm when 55.8 g of \(\ce{Zn}\) metal react with excess \(\ce{HCl}\)?
\[\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}\nonumber \]
Solution
Here we have a stoichiometry problem where we need to find the number of moles of H2 produced. Then we can use the ideal gas law, with the given temperature and pressure, to determine the volume of gas produced. First, the number of moles of H2 is calculated:
\[55.8\cancel{g\, Zn}\times \frac{1\cancel{mol\, Zn}}{65.41\cancel{g\, Zn}}\times \dfrac{1\, mol\, H_{2}}{1\cancel{mol\, Zn}}=0.853\, H_{2}\nonumber \]
Now that we know the number of moles of gas, we can use the ideal gas law to determine the volume, given the other conditions:
\[(1.07atm)V=(0.853\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(299\, K)\nonumber \]
All the units cancel except for L, for volume, which means V = 19.6 L
Exercise \(\PageIndex{3}\)
What pressure of \(\ce{HCl}\) is generated if 3.44 g of \(\ce{Cl2}\) are reacted in 4.55 L at 455 K?
\[\ce{H2(g) + Cl2(g) → 2HCl(g)}\nonumber \]
- Answer
-
0.796 atm
Summary
- The ideal gas law relates the four independent physical properties of a gas at any time.
- The ideal gas law can be used in stoichiometry problems with chemical reactions that involve gases.
