Examples of Determining Volume of a Gas at Different Temperatures
Understanding Volume Changes of a Gas with Temperature
The Ideal Gas Law is a vital equation in chemistry that relates pressure, volume, temperature, and the number of moles of a gas. Mathematically, it is represented as:
\[ PV = nRT \]
Where:
- P = Pressure of the gas
- V = Volume of the gas
- n = Number of moles of gas
- R = Ideal gas constant
- T = Temperature in Kelvin
In this article, we will explore three practical examples of determining the volume of a gas at different temperatures while keeping pressure and the number of moles constant.
Example 1: Volume of Oxygen Gas at Room Temperature
In a laboratory, a chemist is conducting an experiment with a sample of oxygen gas. The sample contains 2 moles of oxygen, and the pressure is maintained at 1 atm. The chemist wants to determine the volume of the gas at room temperature (25°C or 298 K).
Using the Ideal Gas Law:
\[ V = \frac{nRT}{P} \]
Substituting the known values:
- n = 2 moles
- R = 0.0821 L·atm/(K·mol)
- T = 298 K
- P = 1 atm
Calculating:
\[ V = \frac{(2)(0.0821)(298)}{1} \approx 49.2 \text{ L} \]
In this case, the volume of the oxygen gas is approximately 49.2 liters at room temperature.
Notes:
- If the temperature were to change to 0°C (273 K), the new volume could be calculated using the same formula, which illustrates the direct relationship between temperature and volume.
Example 2: Volume of Helium in a Balloon at Different Temperatures
Consider a party balloon filled with helium gas. The balloon contains 0.5 moles of helium, and at 20°C (293 K), the pressure inside the balloon is 1 atm. The host wants to know how the volume changes if the temperature drops to 5°C (278 K).
Using the Ideal Gas Law:
\[ V = \frac{nRT}{P} \]
Initially at 20°C:
- n = 0.5 moles
- R = 0.0821 L·atm/(K·mol)
- T = 293 K
- P = 1 atm
Calculating:
\[ V = \frac{(0.5)(0.0821)(293)}{1} \approx 12.1 \text{ L} \]
Now, calculating for 5°C:
- T = 278 K
Calculating:
\[ V = \frac{(0.5)(0.0821)(278)}{1} \approx 11.4 \text{ L} \]
Thus, the volume decreases from approximately 12.1 liters at 20°C to 11.4 liters at 5°C.
Notes:
- This example illustrates Charles’s Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure remains constant.
Example 3: Compressed Gas in a Cylinder
A gas cylinder contains 3 moles of nitrogen gas compressed to a pressure of 2 atm. The initial temperature of the gas is 100°C (373 K), and the technician needs to determine the volume of gas when it is cooled to 25°C (298 K).
Using the Ideal Gas Law:
\[ V = \frac{nRT}{P} \]
Calculating for the initial temperature (100°C):
- n = 3 moles
- R = 0.0821 L·atm/(K·mol)
- T = 373 K
- P = 2 atm
Calculating:
\[ V = \frac{(3)(0.0821)(373)}{2} \approx 38.5 \text{ L} \]
Now, calculating for the final temperature (25°C):
- T = 298 K
Calculating:
\[ V = \frac{(3)(0.0821)(298)}{2} \approx 36.7 \text{ L} \]
In this scenario, the volume of nitrogen gas decreases from approximately 38.5 liters at 100°C to 36.7 liters at 25°C.
Notes:
- This example highlights how cooling a gas reduces its volume when pressure is held constant. It’s essential for safety in industrial applications where gas volumes can be critical.
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Ideal Gas Law Applications: Gas Mixtures
Ideal Gas Law in Stoichiometry: 3 Examples
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