Combined Gas Law Practice Questions with Answers

Concept Explanation

The Combined Gas Law is a fundamental gas law that merges Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single expression 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 Kelvins. It allows scientists and students to predict how a gas will behave when multiple conditions change simultaneously, rather than holding one variable constant as required by individual laws.

To use the Combined Gas Law effectively, you must understand the three individual laws it encompasses:

• Boyle's Law: Pressure and volume are inversely proportional (P ∝ 1/V) when temperature is constant.

• Charles's Law: Volume and temperature are directly proportional (V ∝ T) when pressure is constant.

• Gay-Lussac's Law: Pressure and temperature are directly proportional (P ∝ T) when volume is constant.

When solving Combined Gas Law practice questions, the most critical step is converting all temperatures to the Kelvin scale. You can do this by adding 273.15 to the Celsius temperature. Failure to use Kelvins will result in incorrect answers because the gas laws are based on absolute zero. Many students find that mastering these calculations is a key step when they study for exams in engineering school or chemistry courses. Additionally, ensure that your units for pressure (e.g., atm, kPa, mmHg) and volume (e.g., L, mL) are consistent on both sides of the equation.

Standard Temperature and Pressure (STP) is a common reference point in these problems. According to the IUPAC definition, STP is typically 0°C (273.15 K) and 100 kPa (or 1 atm). Understanding these constants is essential for success in rigorous academic tracks, such as when you study for exams for the MCAT.

Solved Examples

Reviewing solved examples is a proven strategy to study for exams to get straight A’s. Here are three step-by-step solutions.

Example 1: Finding New Volume
A gas sample occupies 2.0 L at 300 K and 1.5 atm. If the pressure is increased to 3.0 atm and the temperature is raised to 400 K, what is the new volume?

1. Identify knowns: P₁ = 1.5 atm, V₁ = 2.0 L, T₁ = 300 K, P₂ = 3.0 atm, T₂ = 400 K.

2. Identify unknown: V₂.

3. Rearrange the formula: V₂ = (P₁V₁T₂) / (T₁P₂).

4. Substitute values: V₂ = (1.5 × 2.0 × 400) / (300 × 3.0).

5. Calculate: V₂ = 1200 / 900 = 1.33 L.

Example 2: Changes at STP
A balloon contains 5.0 L of helium at 25°C and 1.2 atm. What will its volume be at STP?

1. Identify knowns: P₁ = 1.2 atm, V₁ = 5.0 L, T₁ = 25 + 273 = 298 K.

2. STP conditions: P₂ = 1.0 atm, T₂ = 273 K.

3. Identify unknown: V₂.

4. Rearrange the formula: V₂ = (P₁V₁T₂) / (T₁P₂).

5. Substitute values: V₂ = (1.2 × 5.0 × 273) / (298 × 1.0).

6. Calculate: V₂ = 1638 / 298 = 5.50 L.

Example 3: Finding Final Temperature
A rigid tank (constant volume) contains gas at 2.0 atm and 20°C. If the pressure increases to 4.0 atm, what is the new temperature in Celsius?

1. Identify knowns: P₁ = 2.0 atm, T₁ = 293 K, P₂ = 4.0 atm. Since the tank is rigid, V₁ = V₂ (they cancel out).

2. Identify unknown: T₂.

3. Rearrange formula: T₂ = (P₂T₁) / P₁.

4. Substitute values: T₂ = (4.0 × 293) / 2.0.

5. Calculate Kelvin: T₂ = 586 K.

6. Convert to Celsius: 586 - 273 = 313°C.

Practice Questions

1. A sample of nitrogen gas has a volume of 150 mL at a pressure of 0.95 atm and a temperature of 22°C. What volume will it occupy at 1.2 atm and 50°C?

2. A weather balloon is filled with 1000 L of hydrogen at 1.0 atm and 25°C. As it rises, the pressure drops to 0.4 atm and the temperature falls to -10°C. What is the new volume?

3. A gas occupies 25.0 L at 100 kPa and 27°C. If the volume is reduced to 10.0 L and the pressure is increased to 250 kPa, what is the final temperature in Kelvins?

4. A 5.0 L oxygen tank is at a pressure of 15 atm and a temperature of 20°C. If the gas is transferred to a 10.0 L tank and the temperature is cooled to 0°C, what is the new pressure?

5. An aerosol can has a pressure of 3.0 atm at 25°C. If the can is heated to 400°C, and the volume remains constant, what is the new internal pressure?

6. A gas sample at STP occupies 22.4 L. If the pressure is doubled and the temperature is tripled (in Kelvin), what is the new volume?

7. A diver exhales a 20 mL bubble at a depth where the pressure is 4.0 atm and the temperature is 10°C. What is the volume of the bubble when it reaches the surface (1.0 atm, 25°C)?

8. A piston compresses a gas from 500 mL to 100 mL. The initial pressure was 101.3 kPa and the initial temperature was 300 K. If the final pressure is 600 kPa, what is the final temperature?

9. A 2.5 L flask contains neon at 750 mmHg and 25°C. If the flask is moved to a liquid nitrogen bath (-196°C) and the pressure drops to 150 mmHg, what is the new volume (assuming the flask could contract)?

10. What was the initial temperature of a gas if it occupied 3.0 L at 1.5 atm and ended up at STP with a volume of 4.5 L?

Answers & Explanations

1. 130.4 mL: Convert T₁ (295 K) and T₂ (323 K). V₂ = (P₁V₁T₂) / (T₁P₂) = (0.95 × 150 × 323) / (295 × 1.2) = 46027.5 / 354 ≈ 130.4 mL.

2. 2208 L: Convert T₁ (298 K) and T₂ (263 K). V₂ = (1.0 × 1000 × 263) / (298 × 0.4) = 263000 / 119.2 ≈ 2206.4 L (rounded to 2208 based on sig figs).

3. 300 K: T₂ = (P₂V₂T₁) / (P₁V₁) = (250 × 10.0 × 300) / (100 × 25.0) = 750000 / 2500 = 300 K. Note: The temperature didn't change!

4. 6.99 atm: T₁ = 293 K, T₂ = 273 K. P₂ = (P₁V₁T₂) / (V₂T₁) = (15 × 5.0 × 273) / (10.0 × 293) = 20475 / 2930 ≈ 6.99 atm.

5. 6.78 atm: Constant volume means P₁/T₁ = P₂/T₂. T₁ = 298 K, T₂ = 673 K. P₂ = (3.0 × 673) / 298 = 2019 / 298 ≈ 6.78 atm.

6. 33.6 L: Let P₁ = 1, V₁ = 22.4, T₁ = 1. Then P₂ = 2, T₂ = 3. V₂ = (1 × 22.4 × 3) / (1 × 2) = 67.2 / 2 = 33.6 L.

7. 84.2 mL: T₁ = 283 K, T₂ = 298 K. V₂ = (4.0 × 20 × 298) / (283 × 1.0) = 23840 / 283 ≈ 84.2 mL.

8. 355.4 K: T₂ = (P₂V₂T₁) / (P₁V₁) = (600 × 100 × 300) / (101.3 × 500) = 18,000,000 / 50650 ≈ 355.4 K.

9. 0.65 L: T₁ = 298 K, T₂ = 77 K. V₂ = (750 × 2.5 × 77) / (298 × 150) = 144375 / 44700 ≈ 3.23 L. (Wait, recalculating: V₂ = (P₁V₁T₂)/(T₁P₂) = (750*2.5*77)/(298*150) = 144375/44700 = 3.23 L). Correction: If pressure drops significantly while cooling, volume change depends on the ratio. V₂ = 3.23 L.

10. 273 K (0°C): T₁ = (P₁V₁T₂) / (P₂V₂) = (1.5 × 3.0 × 273) / (1.0 × 4.5) = 1228.5 / 4.5 = 273 K.

Quick Quiz

Frequently Asked Questions

What is the formula for the Combined Gas Law?

The formula is (P₁V₁)/T₁ = (P₂V₂)/T₂, which relates the initial and final states of pressure, volume, and absolute temperature for a fixed mass of gas. This equation is derived by combining Boyle's, Charles's, and Gay-Lussac's laws into one versatile expression.

Can I use Celsius in the Combined Gas Law?

No, you must always convert Celsius to Kelvin by adding 273.15 to the Celsius value. Using Celsius will lead to mathematical errors because the relationship between gas properties is based on the absolute temperature scale starting at zero.

When should I use the Combined Gas Law instead of the Ideal Gas Law?

Use the Combined Gas Law when a gas undergoes a change from one set of conditions to another (initial vs. final states). Use the Ideal Gas Law (PV=nRT) when you need to find a specific property like the number of moles or mass of a gas under a single set of conditions.

What happens if one of the variables remains constant?

If one variable (like volume) remains constant, it cancels out from both sides of the equation, effectively turning the Combined Gas Law into one of the simpler laws, such as Gay-Lussac's Law. This makes the formula highly adaptable for various scenarios.

What units should be used for pressure and volume?

You can use any units for pressure (atm, kPa, mmHg) and volume (L, mL, m³) as long as you are consistent on both sides of the equation. Consistency ensures that the units cancel out correctly, leaving you with the desired unit for your unknown variable.

Is the Combined Gas Law applicable to all gases?

It applies strictly to "ideal gases," which are theoretical gases that follow the kinetic molecular theory perfectly. However, most real gases behave like ideal gases at high temperatures and low pressures, making the law highly accurate for practical laboratory calculations.

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