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Learning Objectives
- Learn and apply the ideal gas law.
- Learn and apply the combined gas law.
There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.
Ideal Gas Law
As with the other gas laws, we can also say that \(\frac{\left( P \times V \right)}{\left( T \times n \right)}\) is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable \(R\) for the constant, the equation becomes:
\[\dfrac{P \times V}{T \times n} = R \nonumber \]
The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:
\[PV = nRT \nonumber \]
The variable \(R\) in the equation is called the ideal gas constant.
Evaluating the Ideal Gas Constant
The value of \(R\), the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: \(\text{kPa}\), \(\text{atm}\), or \(\text{mm} \: \ce{Hg}\). Therefore, \(R\) can have three different values.
We will demonstrate how \(R\) is calculated when the pressure is measured in \(\text{kPa}\). The volume of \(1.00 \: \text{mol}\) of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm)is measured to be \(22.414 \: \text{L}\). We can substitute \(101.325 \: \text{kPa}\) for pressure, \(22.414 \: \text{L}\) for volume, and \(273.15 \: \text{K}\) for temperature into the ideal gas equation and solve for \(R\).
\[\begin{align*} R &= \frac{PV}{nT} \\[4pt] &= \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} \\[4pt] &= 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol} \end{align*} \nonumber \]
This is the value of \(R\) that is to be used in the ideal gas equation when the pressure is given in \(\text{kPa}\). The table below shows a summary of this and the other possible values of \(R\). It is important to choose the correct value of \(R\) to use for a given problem.
| Unit of \(P\) | Unit of \(V\) | Unit of \(n\) | Unit of \(T\) | Value and Unit of \(R\) |
|---|---|---|---|---|
| \(\text{kPa}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(8.314 \: \text{J/K} \cdot \text{mol}\) |
| \(\text{atm}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}\) |
| \(\text{mm} \: \ce{Hg}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}\) |
Notice that the unit for \(R\) when the pressure is in \(\text{kPa}\) has been changed to \(\text{J/K} \cdot \text{mol}\). A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule \(\left( \text{J} \right)\).
Example \(\PageIndex{1}\) Oxygen Gas
What volume is occupied by \(3.76 \: \text{g}\) of oxygen gas at a pressure of \(88.4 \: \text{kPa}\) and a temperature of \(19^\text{o} \text{C}\)? Assume the oxygen is ideal.
Solution
Steps for Problem Solving | Example \(\PageIndex{1}\) |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." | Given:
Mass \(\ce{O_2} = 3.76 \: \text{g}\) Find: V = ? L |
| List other known quantities | \(\ce{O_2} = 32.00 \: \text{g/mol}\) \(R = 8.314 \: \text{J/K} \cdot \text{mol}\) |
| Plan the problem |
\[V = \frac{nRT}{P} \nonumber \] |
| Calculate | 1. \[3.76 \: \cancel{\text{g}} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \cancel{\text{g}} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2} \nonumber \] 2. Now substitute the known quantities into the equation and solve. \[V = \frac{nRT}{P} = \frac{0.1175 \: \cancel{\text{mol}} \times 8.314 \: \cancel{\text{J/K}} \cdot \cancel{\text{mol}} \times 292 \: \cancel{\text{K}}}{88.4 \: \cancel{\text{kPa}}} = 3.23 \: \text{L} \: \ce{O_2} \nonumber \] |
| Think about your result. | The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume \(\left( 22.4 \: \text{L/mol} \right)\) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for \(T\) and \(P\). Since a joule \(\left( \text{J} \right) = \text{kPa} \cdot \text{L}\), the units cancel out correctly, leaving a volume in liters. |
Example \(\PageIndex{2}\): Argon Gas
A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?
Solution
Steps for Problem Solving | Example \(\PageIndex{2}\) |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." | Given: n = 4.22 mol P = 1.21 atm T = 34°C Find: V = ? L |
| List other known quantities | none |
| Plan the problem | 1. The first step is to convert temperature to kelvin. 2. Then, rearrange the equation algebraically to solve for V \[V = \frac{nRT}{P} \nonumber \] |
| Calculate | 1. 34 + 273 = 307 K 2. Now substitute the known quantities into the equation and solve. \[ \begin{align*} V=\frac{(4.22\, \cancel{mol})(0.08205\frac{L.\cancel{atm}}{\cancel{mol.K}})(307\, \cancel{K)}}{1.21\cancel{atm}} \\[4pt] &= 87.9 \,L \end{align*} \nonumber \] |
| Think about your result. | The number of moles of Ar is large so the expected volume should also be large. |
Exercise \(\PageIndex{1}\)
A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?
Answer
3.94 L
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
One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law, and its mathematical form is
\[\frac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n \nonumber \]
This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in Kelvin.
Example \(\PageIndex{3}\):
A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?
Solution
Steps for Problem Solving | Example \(\PageIndex{3}\) |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." | Given: V1 = 8.33 L, P1 = 1.82 atm, and T1 = 286 K V2 = 5.72 L and T2 = 355 K Find: P2 = ? atm |
| List other known quantities | none |
| Plan the problem | First, rearrange the equation algebraically to solve for \(V_2\). \(P_2 = \frac{P_1 V_1 T_2 }{T_1V_2}\) |
| Calculate | Now substitute the known quantities into the equation and solve. \[P_2 = \frac{(1.82\, atm)(8.33\, \cancel{L})(355\, \cancel{K})}{(286\, \cancel{K})(5.72\, \cancel{L})}=3.22 atm \nonumber \] |
| Think about your result. | Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing. |
Exercise \(\PageIndex{3}\)
If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2?
- Answer
-
105 mL
As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.
Summary
- The ideal gas constant is calculated to be \(8.314 \: \text{J/K} \cdot \text{mol}\) when the pressure is in kPa.
- The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
- The combined gas law relates pressure, volume, and temperature of a gas.
Contributors and Attributions
Henry Agnew(UC Davis)
