Name: Date: Period: Gas Laws Worksheet: Boyle, Charles, and Combined Gas Laws Boyle’s Law Problems: P, VP,V, 1 atm = 760.0 mm Hg = 101.3 kPa=760.0 torr 1. If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature. What is the new volume? 2. A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temperature remains constant? Charles’ Law Problems: atm = 760.0 mm Hg = 101.3 kPa= 760.0 torr 3. Calculate the decrease in temperature when 6.00 Lat 20.0 € is compressed to 4.00 L.
The Correct Answer and Explanation is :
Boyle’s Law Problems
Boyle’s Law states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional. The formula for Boyle’s Law is:
[
P_1 \cdot V_1 = P_2 \cdot V_2
]
Where:
- (P_1) = initial pressure
- (V_1) = initial volume
- (P_2) = final pressure
- (V_2) = final volume
Problem 1:
Given:
- (V_1 = 22.5) L
- (P_1 = 748) mm Hg
- (P_2 = 725) mm Hg
- (V_2 = ?)
Using Boyle’s Law:
[
P_1 \cdot V_1 = P_2 \cdot V_2
]
Substitute the known values:
[
748 \cdot 22.5 = 725 \cdot V_2
]
Solve for (V_2):
[
V_2 = \frac{748 \cdot 22.5}{725} = 23.2 \, \text{L}
]
So, the new volume is 23.2 L.
Problem 2:
Given:
- (V_1 = 4.0) L
- (P_1 = 205) kPa
- (V_2 = 12.0) L
- (P_2 = ?)
Using Boyle’s Law again:
[
P_1 \cdot V_1 = P_2 \cdot V_2
]
Substitute the known values:
[
205 \cdot 4.0 = P_2 \cdot 12.0
]
Solve for (P_2):
[
P_2 = \frac{205 \cdot 4.0}{12.0} = 68.3 \, \text{kPa}
]
So, the final pressure is 68.3 kPa.
Charles’s Law Problems
Charles’s Law states that the volume of a gas is directly proportional to its temperature when the pressure is constant. The formula for Charles’s Law is:
[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
]
Where:
- (V_1) = initial volume
- (T_1) = initial temperature in Kelvin
- (V_2) = final volume
- (T_2) = final temperature in Kelvin
Problem 3:
Given:
- (V_1 = 6.00) L
- (T_1 = 20.0^\circ C = 293.15 \, \text{K}) (converted to Kelvin)
- (V_2 = 4.00) L
- (T_2 = ?)
Rearrange Charles’s Law to solve for (T_2):
[
T_2 = \frac{V_2 \cdot T_1}{V_1}
]
Substitute the known values:
[
T_2 = \frac{4.00 \cdot 293.15}{6.00} = 195.43 \, \text{K}
]
Convert (T_2) back to Celsius:
[
T_2 = 195.43 – 273.15 = -77.7^\circ C
]
So, the final temperature is -77.7°C.
Explanation:
- Boyle’s Law applies when the temperature is constant. As the pressure on a gas increases, the volume decreases, and vice versa. In Problem 1, the volume decreased due to a slight reduction in pressure, so we calculate the new volume using the inverse relationship.
- In Problem 2, the gas expanded, so the volume increased while the pressure decreased.
- Charles’s Law describes how gases expand or contract when the temperature changes, assuming pressure is constant. In Problem 3, as the gas is compressed, the temperature decreases, showing the inverse relationship between temperature and volume.
These laws demonstrate the basic principles of gas behavior and are foundational for understanding thermodynamics and the behavior of gases in different conditions.