Medium Combined Gas Law Practice Questions

Concept Explanation

The combined gas law is a chemistry principle that merges Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single equation to describe the relationship between pressure, volume, and absolute temperature for a fixed amount of gas. This law is mathematically expressed as (P₁V₁)/T₁ = (P₂V₂)/T₂, where P represents pressure, V represents volume, and T represents temperature in Kelvin. It is particularly useful when all three variables change simultaneously, allowing scientists to predict the final state of a gas system based on its initial conditions. Unlike the Ideal Gas Law, which includes the number of moles (n), the combined gas law assumes the amount of gas remains constant throughout the process.

To use this law effectively, you must adhere to specific units and rules. Temperature must always be converted to the Kelvin scale by adding 273.15 to the Celsius value. While pressure and volume units can vary (such as liters, milliliters, atm, or kPa), they must remain consistent on both sides of the equation. This law is a cornerstone of gas phase stoichiometry and is frequently used in meteorology, scuba diving calculations, and automotive engineering.

Solved Examples

Reviewing these worked examples will help you master the algebraic manipulation required for medium-difficulty problems.

1. Example 1: Finding Final Volume
   A gas sample occupies 2.50 L at 1.20 atm and 25.0°C. What will be the volume if the pressure is increased to 2.50 atm and the temperature is raised to 100.0°C?
   
   1. Identify the knowns: P₁ = 1.20 atm, V₁ = 2.50 L, T₁ = 25.0 + 273 = 298 K, P₂ = 2.50 atm, T₂ = 100.0 + 273 = 373 K.
   2. Rearrange the formula for V₂: V₂ = (P₁V₁T₂) / (T₁P₂).
   3. Substitute the values: V₂ = (1.20 atm × 2.50 L × 373 K) / (298 K × 2.50 atm).
   4. Calculate the result: V₂ = 1119 / 745 = 1.50 L.
2. Example 2: Finding Final Temperature
   A weather balloon contains 50.0 L of helium at 740 mmHg and 20.0°C. If the balloon rises to an altitude where the pressure is 350 mmHg and the volume expands to 90.0 L, what is the new temperature in Celsius?
   
   1. Identify the knowns: P₁ = 740 mmHg, V₁ = 50.0 L, T₁ = 293 K, P₂ = 350 mmHg, V₂ = 90.0 L.
   2. Rearrange for T₂: T₂ = (P₂V₂T₁) / (P₁V₁).
   3. Substitute: T₂ = (350 mmHg × 90.0 L × 293 K) / (740 mmHg × 50.0 L).
   4. Calculate Kelvin: T₂ = 9,229,500 / 37,000 = 249.4 K.
   5. Convert to Celsius: 249.4 - 273 = -23.6°C.
3. Example 3: Standard Temperature and Pressure (STP)
   A gas has a volume of 4.0 L at 2.0 atm and 300 K. What is its volume at STP?
   
   1. Identify knowns: P₁ = 2.0 atm, V₁ = 4.0 L, T₁ = 300 K.
   2. Identify STP conditions: P₂ = 1.0 atm, T₂ = 273 K.
   3. Rearrange for V₂: V₂ = (P₁V₁T₂) / (T₁P₂).
   4. Substitute: V₂ = (2.0 × 4.0 × 273) / (300 × 1.0).
   5. Calculate: V₂ = 2184 / 300 = 7.28 L.

Practice Questions

Test your understanding with these medium-level problems. Remember to keep your units consistent and always convert to Kelvin.

1. A sample of nitrogen gas is stored in a 5.00 L container at 3.00 atm and 27.0°C. If the gas is transferred to a 10.0 L container and the pressure drops to 1.20 atm, what is the new temperature of the gas in Kelvin?

2. A cylinder with a moveable piston contains 250 mL of air at 10.0°C and 1.50 atm. If the volume is compressed to 100 mL and the pressure increases to 4.50 atm, what is the final temperature in degrees Celsius?

3. A gas occupies 12.0 L at 0.80 atm and 200 K. If the temperature is increased to 400 K and the volume is reduced to 6.0 L, what will be the new pressure?

4. An oxygen tank has a volume of 15.0 L at a pressure of 150 atm and a temperature of 25°C. The oxygen is released into a room at 1.00 atm and 22°C. What volume does the released oxygen occupy?

5. A balloon filled with 2.0 L of helium at 1.0 atm and 300 K is taken underwater. The new pressure is 3.5 atm and the temperature is 280 K. What is the new volume of the balloon?

6. A gas sample at 45.0°C and 1.10 atm occupies 3.45 L. What pressure is required to reduce the volume to 2.00 L at a temperature of 15.0°C?

7. A 2.50 L sample of gas at STP is heated to 500 K and the pressure is increased to 3.20 atm. Calculate the new volume.

8. A flexible container holds 45.0 L of gas at 95 kPa and 25°C. If the pressure is increased to 150 kPa and the volume is decreased to 30.0 L, what is the final temperature in Kelvin?

9. A gas at 1.5 atm and 298 K occupies 10.0 L. If the volume stays constant but the temperature drops to 273 K, what is the new pressure? (Note: This is a specific case of the combined law, similar to Boyle's Law variations).

10. If a gas at 700 mmHg and 30°C occupies 25.0 L, what will be its volume at 800 mmHg and 10°C?

Answers & Explanations

Use the following explanations to check your work and understand the logic behind each solution.

• 1. Answer: 240 K
  Using (P₁V₁)/T₁ = (P₂V₂)/T₂: (3.00 × 5.00) / 300 = (1.20 × 10.0) / T₂. 15 / 300 = 12 / T₂. 0.05 = 12 / T₂. T₂ = 12 / 0.05 = 240 K.
• 2. Answer: 67.8°C
  Knowns: P₁=1.50, V₁=250, T₁=283, P₂=4.50, V₂=100. T₂ = (4.50 × 100 × 283) / (1.50 × 250) = 127,350 / 375 = 340.8 K. Celsius = 340.8 - 273 = 67.8°C.
• 3. Answer: 3.2 atm
  P₂ = (P₁V₁T₂) / (T₁V₂) = (0.80 × 12.0 × 400) / (200 × 6.0) = 3840 / 1200 = 3.2 atm.
• 4. Answer: 2227.3 L
  V₂ = (150 × 15.0 × 295) / (298 × 1.00). V₂ = 663,750 / 298 = 2227.3 L. (Note: Units for P and V are consistent).
• 5. Answer: 0.53 L
  V₂ = (1.0 × 2.0 × 280) / (300 × 3.5) = 560 / 1050 = 0.533 L.
• 6. Answer: 1.72 atm
  T₁=318 K, T₂=288 K. P₂ = (1.10 × 3.45 × 288) / (318 × 2.00) = 1092.96 / 636 = 1.718 atm.
• 7. Answer: 1.43 L
  STP means P₁=1.0 atm, T₁=273 K. V₂ = (1.0 × 2.50 × 500) / (273 × 3.20) = 1250 / 873.6 = 1.43 L.
• 8. Answer: 313.7 K
  T₂ = (150 × 30.0 × 298) / (95 × 45.0) = 1,341,000 / 4275 = 313.7 K.
• 9. Answer: 1.37 atm
  P₂ = (1.5 × 10.0 × 273) / (298 × 10.0) = 4095 / 2980 = 1.374 atm. This demonstrates that when volume is constant, P and T are directly proportional.
• 10. Answer: 20.5 L
  V₂ = (700 × 25.0 × 283) / (303 × 800) = 4,952,500 / 242,400 = 20.43 L.

Quick Quiz

Frequently Asked Questions

What is the combined gas law formula?

The formula is (P₁V₁)/T₁ = (P₂V₂)/T₂, where P is pressure, V is volume, and T is temperature in Kelvin. It combines the relationships found in Boyle’s, Charles’s, and Gay-Lussac’s laws.

Can I use Celsius in combined gas law calculations?

No, you must always use Kelvin for temperature because gas laws are based on absolute temperature scales. Using Celsius will lead to incorrect ratios and impossible results when dealing with negative temperatures.

What units should I use for pressure and volume?

You can use any units for pressure (atm, kPa, mmHg) and volume (L, mL) as long as they are consistent for both the initial and final states. If P₁ is in atm, P₂ must also be in atm.

When should I use the combined gas law instead of the ideal gas law?

Use the combined gas law when you are comparing two different states of the same gas sample where the amount of gas (moles) does not change. Use the Ideal Gas Law when you need to find the number of moles or when you only know one set of conditions.

How do I handle STP in these problems?

Treat STP as a set of given values: 0°C (273.15 K) and 1 atm (or 101.325 kPa). These values serve as your P₂ and T₂ (or P₁ and T₁) in the combined gas law equation.

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