The Ideal Gas Law
This is part of Year 11 HSC Chemistry course under the topic of Gas Laws.
HSC Chemistry Syllabus
- Conduct investigations and solve problems to determine the relationship between the Ideal Gas Law and:
- Gay-Lussac's Law (Temperature)
- Boyle's Law
- Charles' Law
- Avogadro's Law
The Combined Gas Law
The Combined Gas Law unites Boyle's, Charles's, and Gay-Lussac's laws into a single expression:
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
Using the Combined Gas Law:
Question 1: Given an initial volume of 8.33L, pressure of 1.82atm, and temperature of 286K, which changes to a temperature of 355K and a volume of 5.72L, what is the final pressure?
Solution:
$$ \text{Identify:}$$
$$ V_1 = 8.33\, \text{L}, P_1 = 1.82\, \text{atm}, T_1 = 286\, \text{K}$$
$$V_2 = 5.72\, \text{L}, P_2 = ?\, \text{atm}, T_2 = 355\, \text{K}$$
$$\text{Substitute and Solve: }$$
$$\frac{P_1 \times V_1}{T_1} \div \frac{V_2}{T_2} = P_2$$
$$P_2 = \frac{(1.82\, \text{atm} \times 8.33\, \text{L})}{286\, \text{K}} \div \frac{5.72\, \text{L}}{355\, \text{K}}$$
$$P_2 = 3.29\, \text{atm}$$
The Ideal Gas Law
The Ideal Gas Law combines the relationships described by Boyle's law, Charles's law, and Avogadro's law. These laws explain how the various properties of gases affect volume:
- Boyle's law on pressure: `V \propto \frac{1}{P}`
- Charles's law on temperature: `V \propto T`
- Avogadro's Law on amounts: `V \propto n`
Combining these individual effects, we obtain the equation known as the ideal gas law:
`V \propto \frac{nT}{P}` or `PV \propto nT` or `\frac{PV}{nT} = R`
Here `R` represents a proportionality constant known as the universal gas constant. The most common form of the ideal gas equation is
$$PV = nRT$$
Where `R = 8.314 J mol^-1K^-1`
Solving Gas Law Problems
To solve problems involving the gas laws, follow these steps:
- Summarize the changing gas variables – knowns and unknowns – and those held constant.
- Convert units, if necessary.
- Rearrange the ideal gas law to obtain the needed relationship of variables and solve for the unknown.
Example:
Question: A gas sample has an initial volume of 1.00 L, an initial pressure of 1.00 atm, and an initial temperature of 273 K. If the volume is increased to 2.00 L and the temperature is raised to 546 K, what is the final pressure, assuming the number of moles of gas remains constant?
Solution:
- Identify the given information and what you need to find:
$$V_1 = 1.00 L, P_1 = 1.00 atm, T_1 = 273 K$$
$$V_2 = 2.00 L, T_2 = 546 K, P_2 = ?$$
- Convert units if necessary: No conversion needed
- Use the combined gas law
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
$$\frac{1.00 \times 1.00}{273} = \frac{P_2 \times 200}{546}$$
$$P_2 = \frac{1.00 \times 1.00 \times 546}{273 \times 2.00} = 1.00 \, \text{atm}$$
The final pressure of the gas is 1.00 atm