Medium Charles’s Law Practice Questions

Concept Explanation

Charles’s Law states that the volume of a given mass of gas is directly proportional to its absolute temperature, provided the pressure remains constant. This fundamental principle of thermodynamics implies that as a gas heats up, its particles move faster and push outward, increasing the volume; conversely, cooling the gas causes it to occupy less space. The mathematical expression for Charles’s Law is V₁/T₁ = V₂/T₂, where V represents volume and T represents temperature in Kelvin. It is critical to remember that all gas law calculations must use the Kelvin scale, which you can calculate by adding 273.15 to the Celsius temperature. This relationship is a cornerstone of the kinetic molecular theory, which is often studied alongside other gas behaviors like Boyle’s Law and the Ideal Gas Law. Understanding this law helps explain everything from how hot air balloons rise to why a basketball might seem flat on a cold winter morning. For more context on how these variables interact in more complex scenarios, you might also explore Combined Gas Law practice questions.

Solved Examples

Reviewing worked problems is one of the best ways to master Medium Charles’s Law Practice Questions. Here are three examples involving unit conversions and algebraic manipulation.

Example 1: Finding Final Volume
A sample of nitrogen gas occupies a volume of 450 mL at 25°C. What will be its volume if the temperature is increased to 75°C at constant pressure?

1. Convert temperatures to Kelvin: T₁ = 25 + 273 = 298 K; T₂ = 75 + 273 = 348 K.
2. Identify knowns: V₁ = 450 mL.
3. Set up the equation: V₂ = (V₁ × T₂) / T₁.
4. Calculate: V₂ = (450 mL × 348 K) / 298 K = 525.5 mL.

Example 2: Finding Initial Temperature
A balloon has a volume of 2.5 L at an unknown temperature. When cooled to 10°C, the volume shrinks to 2.1 L. What was the initial temperature in Celsius?

1. Convert the final temperature to Kelvin: T₂ = 10 + 273 = 283 K.
2. Set up the equation to solve for T₁: T₁ = (V₁ × T₂) / V₂.
3. Calculate T₁: (2.5 L × 283 K) / 2.1 L = 336.9 K.
4. Convert back to Celsius: 336.9 - 273 = 63.9°C.

Example 3: Volume Change in Liters
A 0.50 m³ container of oxygen at 300 K is heated until the volume reaches 0.75 m³. What is the new temperature in Kelvin?

1. Identify knowns: V₁ = 0.50 m³, T₁ = 300 K, V₂ = 0.75 m³.
2. Rearrange the formula for T₂: T₂ = (V₂ × T₁) / V₁.
3. Calculate: T₂ = (0.75 m³ × 300 K) / 0.50 m³ = 450 K.

Practice Questions

Test your knowledge with these Medium Charles’s Law Practice Questions. Ensure you convert all temperatures to Kelvin before solving.

1. A gas occupies 1.2 liters at 20°C. If the pressure remains constant, what volume will it occupy at 100°C?

2. A container holds 500 cm³ of helium at 273 K. To what temperature (in Celsius) must the gas be heated to expand the volume to 750 cm³?

3. If a gas sample has a volume of 3.5 L at -50°C, what will its volume be at 50°C?

4. A flexible container contains 15.0 L of gas at 298 K. If the gas is cooled to 250 K, what is the new volume?

5. A sample of neon gas at 50°C occupies 4.2 L. At what temperature (in Kelvin) will the volume be 3.0 L?

6. A syringe contains 60.0 mL of air at 22°C. If the syringe is placed in a boiling water bath (100°C), what is the new volume of the air?

7. A weather balloon is filled with 100 m³ of hydrogen at 15°C. As it rises, it reaches a layer of the atmosphere where the temperature is -40°C. Assuming pressure is constant, what is the new volume?

8. A gas occupies 250 mL at 30°C. If the volume is doubled while pressure stays constant, what is the final temperature in Celsius?

9. A 2.0 L piston is heated from 25°C to 150°C. By how many liters does the volume increase?

10. An experiment requires a gas to occupy exactly 1.0 L. If the gas currently occupies 0.85 L at 273 K, what temperature (in Kelvin) is needed?

Answers & Explanations

1. 1.53 L. First, convert Celsius to Kelvin: T₁ = 293 K, T₂ = 373 K. Use V₂ = (V₁ × T₂) / T₁: (1.2 L × 373 K) / 293 K = 1.527 L.
2. 136.5°C. Solve for T₂: T₂ = (V₂ × T₁) / V₁ = (750 cm³ × 273 K) / 500 cm³ = 409.5 K. Subtract 273 to get 136.5°C.
3. 5.07 L. Convert temperatures: T₁ = 223 K, T₂ = 323 K. V₂ = (3.5 L × 323 K) / 223 K = 5.067 L.
4. 12.58 L. V₂ = (V₁ × T₂) / T₁ = (15.0 L × 250 K) / 298 K = 12.58 L.
5. 230.7 K. Convert T₁ to 323 K. Solve for T₂: T₂ = (V₂ × T₁) / V₁ = (3.0 L × 323 K) / 4.2 L = 230.7 K.
6. 75.85 mL. T₁ = 295 K, T₂ = 373 K. V₂ = (60.0 mL × 373 K) / 295 K = 75.85 mL.
7. 80.9 m³. T₁ = 288 K, T₂ = 233 K. V₂ = (100 m³ × 233 K) / 288 K = 80.9 m³.
8. 333°C. If volume doubles, V₂ = 500 mL. T₁ = 303 K. T₂ = (500 mL × 303 K) / 250 mL = 606 K. 606 - 273 = 333°C.
9. 0.84 L increase. T₁ = 298 K, T₂ = 423 K. V₂ = (2.0 L × 423 K) / 298 K = 2.84 L. The increase is 2.84 - 2.0 = 0.84 L.
10. 321.2 K. T₂ = (V₂ × T₁) / V₁ = (1.0 L × 273 K) / 0.85 L = 321.18 K.

Quick Quiz

Frequently Asked Questions

What is the relationship between volume and temperature in Charles’s Law?

The relationship is directly proportional, meaning that as the temperature of a gas increases, its volume increases at the same rate, provided pressure is constant. This occurs because higher temperatures increase the kinetic energy of gas molecules, causing them to strike container walls more forcefully and push them outward.

Can I use Celsius in Charles’s Law calculations?

No, you must always use the Kelvin scale for Charles’s Law calculations to ensure the proportionality holds true. Using Celsius would result in mathematical errors because the Celsius scale is not an absolute scale and can include negative numbers or zero, which would imply zero or negative volume.

What happens to a gas at Absolute Zero according to Charles’s Law?

Theoretically, at absolute zero (0 K), the volume of an ideal gas would become zero because molecular motion stops entirely. In reality, gases liquefy or solidify before reaching this temperature, as explained in resources from Encyclopedia Britannica.

How does Charles’s Law differ from Boyle’s Law?

Charles’s Law describes the direct relationship between volume and temperature at constant pressure, whereas Boyle’s Law describes the inverse relationship between volume and pressure at constant temperature. Both are specific cases of the more general Ideal Gas Law equation.

Why do hot air balloons rise according to this law?

When the air inside a balloon is heated, its volume increases and its density decreases relative to the cooler air outside. This buoyancy, caused by the expansion described by Charles’s Law, allows the balloon to lift off the ground.

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