11.3: Ideal Gas Law
- Page ID
- 52250
Skills to Develop
- Solve problems using the ideal gas law, \(PV = nRT\).
Introduction
The individual gas laws and the combined gas law all require that the quantity of gas remain constant. The Universal Gas Law (also sometimes galled the Ideal Gas Law) allows us to make calculations on different quantities of gas as well.
The Universal Gas Law Constant
We have considered four laws that describe the behavior of gases: Boyle's Law, Charles's Law, Avogadro's Law, and Gay-Lussac's Law. These three relationships, which show how the volume of a gas depends on pressure, temperature, and the number of moles of gas, can be combined to form the ideal gas law:
\[PV = nRT\]
Where each variable and its units are:
- \(P =\) pressure \(\left( \text{atm} \right)\)
- \(V =\) volume \(\left( \text{L} \right)\)
- \(n =\) number of moles of gas \(\left( \text{mol} \right)\)
- \(T =\) temperature \(\left( \text{K} \right)\)
- \(R =\) ideal gas constant \(= 0.0821 \: \text{atm} \cdot \text{L/mol} \cdot \text{K}\)
Up to this point in gas law calculations, we haven't worried too much about which unit you use for pressure and volume as long as the units matched. Notice that the gas constant, \(R\), has specific units. Your units of pressure and volume must be in \(\text{atm}\) and \(\text{L}\), respectively, because they must match the appropriate units in the constant, \(R\). Moles, of course, always have the unit moles and temperature must always be Kelvin. You can convert the value of \(R\) into values for any set of units for pressure and volume, if you wanted, but the numerical value of \(R\) would also change.
Example 11.3.1
A sample of nitrogen gas, \(\ce{N_2}\), has a volume of \(5.56 \: \text{L}\) at \(0^\text{o} \text{C}\) and \(1.50 \: \text{atm}\) pressure. How many moles of nitrogen are present in this sample?
Solution:
Step 1: Identify the given information & check units. Temperature units must be in Kelvin. Volume and pressure units must match \(R\).
\(P = 1.50 \: \text{atm}\)
\(V = 5.56 \: \text{L}\)
\(n = ?\)
\(T = 273 \: \text{K}\) (must be in \(\text{K}\))
Step 2: Solve the ideal gas law for the unknown variable.
\[PV = nRT\]
\[\left( 1.50 \: \text{atm} \right) \left( 5.56 \: \text{L} \right) = n \left( 0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \right) \left( 273 \: \text{K} \right)\]
\(n = 0.372 \: \text{mol}\)
Example 11.3.2
\(2.00 \: \text{mol}\) of methane gas, \(\ce{CH_4}\), are placed in a rigid \(500 \: \text{mL}\) container and heated to \(100^\text{o} \text{C}\). What pressure will be exerted by the methane?
Solution:
Step 1: Identify the given information & check units. Temperature must be in Kelvin. Volume and pressure units must match \(R\).
\(P = ?\)
\(V = 500 \: \text{mL} = 0.500 \: \text{L}\)
\(n = 2.00 \: \text{mol}\)
\(T = 100^\text{o} \text{C} = 373 \: \text{K}\)
Step 2: Solve the ideal gas law for the unknown variable.
\[PV = nRT\]
\[P \left( 0.500 \: \text{L} \right) = \left( 2.00 \: \text{mol} \right) \left( 0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \right) \left( 373 \: \text{K} \right)\]
\(P = 122 \: \text{atm}\)
Lesson Summary
- The Universal Gas Law: \(PV = nRT\)
- At STP, one mole of any gas occupies \(22.4 \: \text{L}\).
- The universal gas law is often used along with laboratory data to find the molar mass of an unknown substance.