The Combined Gas Law combines three individual gas laws: Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law. It describes the relationship between pressure, volume, and temperature of a gas. The formula is expressed as:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
where:
This law is particularly useful in a variety of practical situations, such as predicting how gases behave under different conditions. Below are three diverse examples to illustrate its application.
Imagine you’re using a syringe to draw up a liquid. Initially, the syringe holds 20 mL of air at a pressure of 1 atm and a temperature of 298 K. You then push the plunger to reduce the volume of the air to 10 mL. What is the new pressure of the air inside the syringe?
In this case, we can use the Combined Gas Law to solve for the new pressure.
Substituting into the formula:
\[ \frac{1 \, \text{atm} \times 20 \, \text{mL}}{298 \, \text{K}} = \frac{P_2 \times 10 \, \text{mL}}{298 \, \text{K}} \]
This simplifies to:
\[ P_2 = \frac{1 \, \text{atm} \times 20 \, \text{mL}}{10 \, \text{mL}} = 2 \, \text{atm} \]
Thus, the new pressure in the syringe is 2 atm.
Note: This example illustrates how gas volume reduction results in increased pressure, demonstrating Boyle’s Law in action within the Combined Gas Law framework.
Consider a balloon that has a volume of 5 L at a pressure of 1 atm and a temperature of 273 K (0°C). If the temperature of the air inside the balloon is increased to 313 K (40°C), what will be its new volume?
In this situation, we will again use the Combined Gas Law to find the new volume.
Using the Combined Gas Law:
\[ \frac{1 \, \text{atm} \times 5 \, \text{L}}{273 \, \text{K}} = \frac{P_2 \times V_2}{313 \, \text{K}} \]
Since \(P_2\) remains constant at 1 atm, we can rearrange to solve for \(V_2\):
\[ V_2 = \frac{1 \, \text{atm} \times 5 \, \text{L} \times 313 \, \text{K}}{1 \, \text{atm} \times 273 \, \text{K}} \approx 5.74 \, \text{L} \]
Therefore, the new volume of the balloon at 40°C is approximately 5.74 L.
Note: This example demonstrates Charles’s Law, where an increase in temperature leads to an increase in volume, assuming pressure is constant.
Imagine a rigid gas container that holds 10 L of gas at a pressure of 2 atm and a temperature of 300 K. If the temperature of the gas is decreased to 250 K, what will be the new pressure?
Here, we will use the Combined Gas Law to find the new pressure while keeping the volume constant.
Using the Combined Gas Law:
\[ \frac{2 \, \text{atm} \times 10 \, \text{L}}{300 \, \text{K}} = \frac{P_2 \times 10 \, \text{L}}{250 \, \text{K}} \]
Simplifying gives us:
\[ P_2 = \frac{2 \, \text{atm} \times 250 \, \text{K}}{300 \, \text{K}} \approx 1.67 \, \text{atm} \]
Thus, the new pressure when the temperature decreases to 250 K is approximately 1.67 atm.
Note: This scenario exemplifies Gay-Lussac’s Law, where a decrease in temperature results in a decrease in pressure when the volume is constant.