Question
Combined gas law formula, relationship, example problem
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Answered by
GPT-4o mini
The Combined Gas Law incorporates Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single equation that relates pressure, volume, and temperature of a gas. The formula for the Combined Gas Law is:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume of the gas,
- \( T \) is the temperature of the gas (in Kelvin),
- Subscripts 1 and 2 refer to the initial and final states of the gas, respectively.
### Relationships:
- **Boyle's Law** states that pressure and volume are inversely related when temperature is constant (\( P_1 V_1 = P_2 V_2 \)).
- **Charles's Law** states that volume and temperature are directly related when pressure is constant (\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)).
- **Gay-Lussac's Law** states that pressure and temperature are directly related when volume is constant (\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)).
### Example Problem:
Suppose you have a gas at a pressure of 2 atm, a volume of 4 L, and a temperature of 300 K. You heat the gas to 450 K and increase the pressure to 3 atm. What will be the new volume (\( V_2 \)) of the gas?
Given:
- \( P_1 = 2 \, \text{atm} \)
- \( V_1 = 4 \, \text{L} \)
- \( T_1 = 300 \, \text{K} \)
- \( P_2 = 3 \, \text{atm} \)
- \( T_2 = 450 \, \text{K} \)
Plug these values into the Combined Gas Law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Substituting in the known values:
\[
\frac{(2 \, \text{atm}) (4 \, \text{L})}{300 \, \text{K}} = \frac{(3 \, \text{atm}) V_2}{450 \, \text{K}}
\]
Now, calculate the left side:
\[
\frac{8 \, \text{atm} \cdot \text{L}}{300 \, \text{K}} = \frac{3 \, \text{atm} \cdot V_2}{450 \, \text{K}}
\]
Calculating the left side:
\[
\frac{8}{300} = 0.02667 \, \text{atm} \cdot \text{L/K}
\]
Thus, we have:
\[
0.02667 = \frac{3 V_2}{450}
\]
Now, solve for \( V_2 \):
First multiply both sides by 450:
\[
450 \cdot 0.02667 = 3 V_2
\]
This simplifies to:
\[
12 = 3 V_2
\]
Dividing both sides by 3:
\[
V_2 = 4 \, \text{L}
\]
### Conclusion:
The new volume \( V_2 \) is 4 L. In this case, even with changes to pressure and temperature, the volume remains unchanged due to the compensating changes in temperature and pressure.
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume of the gas,
- \( T \) is the temperature of the gas (in Kelvin),
- Subscripts 1 and 2 refer to the initial and final states of the gas, respectively.
### Relationships:
- **Boyle's Law** states that pressure and volume are inversely related when temperature is constant (\( P_1 V_1 = P_2 V_2 \)).
- **Charles's Law** states that volume and temperature are directly related when pressure is constant (\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)).
- **Gay-Lussac's Law** states that pressure and temperature are directly related when volume is constant (\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)).
### Example Problem:
Suppose you have a gas at a pressure of 2 atm, a volume of 4 L, and a temperature of 300 K. You heat the gas to 450 K and increase the pressure to 3 atm. What will be the new volume (\( V_2 \)) of the gas?
Given:
- \( P_1 = 2 \, \text{atm} \)
- \( V_1 = 4 \, \text{L} \)
- \( T_1 = 300 \, \text{K} \)
- \( P_2 = 3 \, \text{atm} \)
- \( T_2 = 450 \, \text{K} \)
Plug these values into the Combined Gas Law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Substituting in the known values:
\[
\frac{(2 \, \text{atm}) (4 \, \text{L})}{300 \, \text{K}} = \frac{(3 \, \text{atm}) V_2}{450 \, \text{K}}
\]
Now, calculate the left side:
\[
\frac{8 \, \text{atm} \cdot \text{L}}{300 \, \text{K}} = \frac{3 \, \text{atm} \cdot V_2}{450 \, \text{K}}
\]
Calculating the left side:
\[
\frac{8}{300} = 0.02667 \, \text{atm} \cdot \text{L/K}
\]
Thus, we have:
\[
0.02667 = \frac{3 V_2}{450}
\]
Now, solve for \( V_2 \):
First multiply both sides by 450:
\[
450 \cdot 0.02667 = 3 V_2
\]
This simplifies to:
\[
12 = 3 V_2
\]
Dividing both sides by 3:
\[
V_2 = 4 \, \text{L}
\]
### Conclusion:
The new volume \( V_2 \) is 4 L. In this case, even with changes to pressure and temperature, the volume remains unchanged due to the compensating changes in temperature and pressure.