📚 What is the Combined Gas Law?
The Combined Gas Law is a fundamental principle in chemistry and physics that relates the pressure, volume, and temperature of a fixed amount of gas. It's particularly useful when dealing with situations where these three properties change simultaneously. It elegantly combines Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single, comprehensive equation.
📜 A Brief History
The Combined Gas Law wasn't discovered by a single person but rather evolved from the work of several scientists over time:
- 💨 Robert Boyle (1662): 🌡️ Established Boyle's Law, which states that at constant temperature, the pressure and volume of a gas are inversely proportional.
- 🔥 Jacques Charles (1780s): 🎈 Discovered Charles's Law, stating that at constant pressure, the volume of a gas is directly proportional to its absolute temperature.
- 🌡️ Joseph Louis Gay-Lussac (1802): 💥 Formulated Gay-Lussac's Law, which states that at constant volume, the pressure of a gas is directly proportional to its absolute temperature.
⚗️ The Key Principles
The Combined Gas Law is expressed mathematically as:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Where:
- 📌 $P_1$ = Initial pressure
- 📐 $V_1$ = Initial volume
- 🌡️ $T_1$ = Initial absolute temperature (in Kelvin)
- 📍 $P_2$ = Final pressure
- 📏 $V_2$ = Final volume
- ♨️ $T_2$ = Final absolute temperature (in Kelvin)
Important Notes:
- 🧊 Temperature must always be in Kelvin (K). To convert from Celsius (°C) to Kelvin, use the formula: $K = °C + 273.15$.
- ⚖️ The amount of gas (number of moles) remains constant.
- 📐 The units for pressure and volume must be consistent on both sides of the equation (e.g., both in atm or both in L).
🌍 Real-World Examples
- 🎈 Weather Balloons: 🛰️ As a weather balloon ascends, the atmospheric pressure decreases, and the temperature changes. The Combined Gas Law helps predict the balloon's volume at different altitudes.
- 🚗 Car Tires: 🌡️ The pressure in a car tire increases after driving due to the increase in temperature. The Combined Gas Law can be used to estimate the pressure change.
- 🤿 Scuba Diving: 🌊 The volume of air in a scuba tank changes with depth due to pressure and temperature variations.
🧪 Sample Problems & Solutions
Problem 1
A gas occupies a volume of 10.0 L at standard temperature and pressure (STP). If the temperature is increased to 100.0°C and the pressure is increased to 2.00 atm, what is the new volume of the gas?
Solution:
First, convert the temperature to Kelvin:
$T_1 = 0°C + 273.15 = 273.15 K$
$T_2 = 100.0°C + 273.15 = 373.15 K$
At STP, $P_1 = 1 atm$ and $V_1 = 10.0 L$. We are given $P_2 = 2.00 atm$.
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(1.00 atm)(10.0 L)}{273.15 K} = \frac{(2.00 atm)(V_2)}{373.15 K}$
Solving for $V_2$:
$V_2 = \frac{(1.00 atm)(10.0 L)(373.15 K)}{(2.00 atm)(273.15 K)} = 6.83 L$
Problem 2
A gas occupies 5.0 L at 25°C and 1.5 atm. What volume will it occupy at 50°C and 3.0 atm?
Solution:
Convert the temperature to Kelvin:
$T_1 = 25°C + 273.15 = 298.15 K$
$T_2 = 50°C + 273.15 = 323.15 K$
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(1.5 atm)(5.0 L)}{298.15 K} = \frac{(3.0 atm)(V_2)}{323.15 K}$
Solving for $V_2$:
$V_2 = \frac{(1.5 atm)(5.0 L)(323.15 K)}{(3.0 atm)(298.15 K)} = 2.71 L$
Problem 3
A container of gas has a volume of 2.0 L at a pressure of 3.0 atm and a temperature of 200 K. If the pressure is decreased to 1.0 atm and the temperature is increased to 300 K, what is the new volume of the gas?
Solution:
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(3.0 atm)(2.0 L)}{200 K} = \frac{(1.0 atm)(V_2)}{300 K}$
Solving for $V_2$:
$V_2 = \frac{(3.0 atm)(2.0 L)(300 K)}{(1.0 atm)(200 K)} = 9.0 L$
Problem 4
A gas occupies 15.0 L at 2.0 atm and 27°C. What volume will it occupy at standard temperature and pressure (STP)?
Solution:
Convert the temperature to Kelvin:
$T_1 = 27°C + 273.15 = 300.15 K$
At STP, $P_2 = 1 atm$ and $T_2 = 273.15 K$.
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(2.0 atm)(15.0 L)}{300.15 K} = \frac{(1.0 atm)(V_2)}{273.15 K}$
Solving for $V_2$:
$V_2 = \frac{(2.0 atm)(15.0 L)(273.15 K)}{(1.0 atm)(300.15 K)} = 27.3 L$
Problem 5
A gas has a volume of 8.0 L at 0.5 atm and -23°C. What is its volume at 1.0 atm and 27°C?
Solution:
Convert the temperatures to Kelvin:
$T_1 = -23°C + 273.15 = 250.15 K$
$T_2 = 27°C + 273.15 = 300.15 K$
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(0.5 atm)(8.0 L)}{250.15 K} = \frac{(1.0 atm)(V_2)}{300.15 K}$
Solving for $V_2$:
$V_2 = \frac{(0.5 atm)(8.0 L)(300.15 K)}{(1.0 atm)(250.15 K)} = 4.8 L$
Problem 6
A gas occupies a volume of 4.0 L at 227°C and 0.8 atm. What is its volume at 27°C and 1.2 atm?
Solution:
Convert the temperatures to Kelvin:
$T_1 = 227°C + 273.15 = 500.15 K$
$T_2 = 27°C + 273.15 = 300.15 K$
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(0.8 atm)(4.0 L)}{500.15 K} = \frac{(1.2 atm)(V_2)}{300.15 K}$
Solving for $V_2$:
$V_2 = \frac{(0.8 atm)(4.0 L)(300.15 K)}{(1.2 atm)(500.15 K)} = 1.6 L$
Problem 7
A gas is held in a 5.0 L container at a temperature of 30°C and a pressure of 2.0 atm. If the temperature is decreased to 0°C and the pressure is increased to 4.0 atm, what is the new volume of the gas?
Solution:
Convert the temperatures to Kelvin:
$T_1 = 30°C + 273.15 = 303.15 K$
$T_2 = 0°C + 273.15 = 273.15 K$
Using the Combined Gas Law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$\frac{(2.0 atm)(5.0 L)}{303.15 K} = \frac{(4.0 atm)(V_2)}{273.15 K}$
Solving for $V_2$:
$V_2 = \frac{(2.0 atm)(5.0 L)(273.15 K)}{(4.0 atm)(303.15 K)} = 2.25 L$
🎓 Conclusion
The Combined Gas Law is a powerful tool for understanding and predicting the behavior of gases under varying conditions. By mastering this law and its applications, you'll gain a deeper understanding of chemistry and physics.