Examples of Gas Law Problems with Solutions: 3 Practical Examples You’ll Actually Use

Before any formulas, let’s anchor this in reality. When teachers talk about examples of gas law problems with solutions: 3 practical examples, they usually mean three workhorse ideas:

• How gas pressure changes when volume changes (Boyle’s Law)
• How gas volume changes with temperature (Charles’s Law)
• How pressure, volume, temperature, and moles all connect (Ideal Gas Law)

Instead of treating them as abstract equations, we’ll use real examples that feel familiar: a syringe, a hot-air balloon, and a scuba tank. Around those three main cases, we’ll layer in several more examples so you see the patterns, not just isolated questions.

Practical Example 1: Boyle’s Law in action (pressure–volume tradeoffs)

Boyle’s Law describes how pressure and volume of a gas relate when temperature and amount of gas stay constant:

\[ P_1 V_1 = P_2 V_2 \]

Where:

• \(P\) is pressure
• \(V\) is volume
• Subscripts 1 and 2 are “before” and “after” states

This is the go-to law when you compress or expand a gas without changing temperature much.

Example 1A: Compressing air in a syringe

This is one of the cleanest examples of gas law problems with solutions because you can feel it with your hands.

Problem
A syringe contains 5.0 mL of air at 1.0 atm. You push the plunger in (no gas escapes) until the volume is 2.0 mL. Assume temperature stays roughly constant. What is the new pressure of the air?

Solution
Use Boyle’s Law: \(P_1 V_1 = P_2 V_2\)

• \(P_1 = 1.0\) atm
• \(V_1 = 5.0\) mL
• \(V_2 = 2.0\) mL
• \(P_2 = ?\)

[
P_1 V_1 = P_2 V_2 \
(1.0\,\text{atm})(5.0\,\text{mL}) = P_2 (2.0\,\text{mL}) \ P_2 = \frac{(1.0)(5.0)}{2.0} = 2.5\,\text{atm}
]

Interpretation
By shrinking the volume to less than half, you increased pressure to 2.5 times atmospheric pressure. This is a classic example of how gas laws explain why syringes push back against your hand.

Example 1B: Scuba diver’s lungs and Boyle’s Law

Here’s a more applied example of Boyle’s Law that matters for safety.

Problem
A scuba diver takes a breath at 66 ft below the surface, where the pressure is about 3.0 atm (rough approximation: every 33 ft adds ~1 atm of pressure on top of the 1 atm at the surface). Her lungs hold 6.0 L of air at that depth. If she rises to the surface and doesn’t exhale, what volume will that air occupy at 1.0 atm, assuming constant temperature?

Solution
Again, \(P_1 V_1 = P_2 V_2\)

• \(P_1 = 3.0\) atm
• \(V_1 = 6.0\) L
• \(P_2 = 1.0\) atm
• \(V_2 = ?\)

[
(3.0\,\text{atm})(6.0\,\text{L}) = (1.0\,\text{atm}) V_2 \ V_2 = 18\,\text{L}
]

Her lung volume would try to expand to 18 L—far more than a human chest can handle—which is why dive training (see resources from NOAA at https://oceanservice.noaa.gov) warns divers never to hold their breath while ascending. This is one of the best examples of gas law problems with solutions that connects textbook math to real-world medical risk.

Example 1C: Weather balloon at constant temperature (short check)

You fill a flexible weather balloon with 2.0 L of helium at 4.0 atm in a lab. It is released and rises until the pressure outside is 1.0 atm. If we pretend temperature stays constant for this quick calculation, what is the new volume?

[
P_1 V_1 = P_2 V_2 \
(4.0)(2.0) = (1.0) V_2 \ V_2 = 8.0\,\text{L}
]

This kind of quick calculation shows up in many examples of gas law problems with solutions: 3 practical examples style worksheets, because it trains you to spot direct pressure–volume relationships.

Practical Example 2: Charles’s Law and temperature–volume changes

Charles’s Law shows how the volume of a gas changes with temperature (in Kelvin) at constant pressure and constant moles:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where temperature must be in Kelvin (K), not Fahrenheit or Celsius.

Example 2A: Hot-air balloon volume and morning vs. afternoon flights

If you’ve ever wondered why balloon pilots love early mornings, here’s an example of how temperature affects gas volume.

Problem
A hot-air balloon’s envelope holds 2800 m³ of air at 68 °F. By midday, the air inside has been heated to 212 °F while the pressure remains approximately constant. What is the new volume of the air if the amount of gas stays the same?

Step 1: Convert temperatures to Kelvin
First convert Fahrenheit to Celsius, then to Kelvin.

For 68 °F:
[
T(°C) = \frac{5}{9}(T(°F) - 32) = \frac{5}{9}(68 - 32) = 20 °C \
T_1 = 20 + 273 = 293\,\text{K}
]

For 212 °F (boiling point of water):
[
T(°C) = \frac{5}{9}(212 - 32) = 100 °C \
T_2 = 100 + 273 = 373\,\text{K}
]

Step 2: Apply Charles’s Law

• \(V_1 = 2800\) m³
• \(T_1 = 293\) K
• \(T_2 = 373\) K
• \(V_2 = ?\)

[
\frac{V_1}{T_1} = \frac{V_2}{T_2} \
\frac{2800}{293} = \frac{V_2}{373} \ V_2 = 2800 \times \frac{373}{293} \approx 3566\,\text{m}^3
]

The heated air expands to about 3566 m³, increasing lift. This is one of the more realistic examples of gas law problems with solutions you’ll see in physics classes, especially when discussing flight and buoyancy.

Example 2B: Car tire pressure on a hot highway

Here’s a quick, modern twist: the National Highway Traffic Safety Administration (NHTSA, https://www.nhtsa.gov) warns that tire blowouts are more common in summer. Gas laws explain why.

Problem
A car tire contains air at 32 psi (gauge) when the temperature is 50 °F in the morning. Later, the air inside warms to 110 °F while the tire’s volume and amount of air stay about the same. Estimate the new pressure, assuming the air behaves ideally.

Step 1: Convert to absolute pressure
Gauge pressure ignores atmospheric pressure. For gas law calculations, use absolute pressure.

Atmospheric pressure ≈ 14.7 psi.

[
P_{1,\text{abs}} = 32 + 14.7 = 46.7\,\text{psi}
]

Step 2: Convert temperatures to Kelvin

For 50 °F:
[
T_1(°C) = \frac{5}{9}(50 - 32) = 10 °C \ T_1 = 283\,\text{K}
]

For 110 °F:
[
T_2(°C) = \frac{5}{9}(110 - 32) \approx 43.3 °C \ T_2 \approx 316\,\text{K}
]

Step 3: Use the proportionality \(P \propto T\)
At constant volume and moles, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\).

[
\frac{46.7}{283} = \frac{P_2}{316} \ P_2 = 46.7 \times \frac{316}{283} \approx 52.1\,\text{psi (absolute)}
]

Convert back to gauge pressure:
[
P_{2,\text{gauge}} = 52.1 - 14.7 \approx 37.4\,\text{psi}
]

That’s a jump of over 5 psi, which is why tire makers and safety agencies emphasize checking tire pressure when tires are cold.

Example 2C: Helium balloon in winter vs. indoors

A helium balloon is filled to 3.0 L at room temperature (72 °F). It’s taken outside into winter air at 23 °F. What happens to the volume, assuming constant pressure?

Convert temperatures:

For 72 °F:
[
T_1(°C) = \frac{5}{9}(72 - 32) \approx 22.2 °C \ T_1 \approx 295\,\text{K}
]

For 23 °F:
[
T_2(°C) = \frac{5}{9}(23 - 32) \approx -5.0 °C \ T_2 \approx 268\,\text{K}
]

Apply Charles’s Law:

[
\frac{3.0}{295} = \frac{V_2}{268} \ V_2 = 3.0 \times \frac{268}{295} \approx 2.7\,\text{L}
]

The balloon shrinks noticeably, a simple example of gas law behavior that students can observe without a lab.

Practical Example 3: Ideal Gas Law as the all-in-one workhorse

The Ideal Gas Law ties together pressure, volume, temperature, and moles:

\[ PV = nRT \]

Where:

• \(P\) = pressure
• \(V\) = volume
• \(n\) = moles of gas
• \(R\) = gas constant (0.0821 L·atm·mol⁻¹·K⁻¹ is a handy version)
• \(T\) = temperature in Kelvin

This is where many of the best examples of gas law problems with solutions: 3 practical examples end up, because the equation is flexible. You can solve for any missing variable if you know the others.

Example 3A: CO₂ in a soda can

Carbonated drinks are a great, modern example of gas laws used in industry.

Problem
A 12.0 oz soda can (about 0.355 L) contains carbon dioxide gas in the headspace at 25 °C and 3.00 atm. How many moles of CO₂ are in the gas phase above the liquid, assuming ideal behavior?

Step 1: Convert temperature to Kelvin

[
T = 25 + 273 = 298\,\text{K}
]

Step 2: Use Ideal Gas Law

• \(P = 3.00\) atm
• \(V = 0.355\) L
• \(T = 298\) K
• \(R = 0.0821\,\text{L·atm·mol}^{-1}\text{·K}^{-1}\)
• \(n = ?\)

[
PV = nRT \
(3.00)(0.355) = n(0.0821)(298) \
1.065 = n(24.48) \
n = \frac{1.065}{24.48} \approx 0.0435\,\text{mol}
]

So there are about 0.044 moles of CO₂ in the gas space. Beverage companies use this kind of calculation constantly, alongside more advanced thermodynamics, to control carbonation levels.

Example 3B: Medical oxygen cylinder for hospital or home care

Medical oxygen is a timely example, especially after the COVID-19 era, where demand for oxygen cylinders increased worldwide. Agencies like the National Institutes of Health (NIH, https://www.nih.gov) and hospitals rely on accurate gas calculations for patient safety.

Problem
A steel oxygen cylinder has a volume of 10.0 L and contains O₂ at 150 atm and 20 °C. Assuming ideal behavior, how many moles of O₂ are in the tank? If a patient uses oxygen at a rate of 2.0 L/min at 1.0 atm and 20 °C, how long will the cylinder last?

Step 1: Moles of O₂ in the cylinder

Convert temperature:

[
T = 20 + 273 = 293\,\text{K}
]

Use Ideal Gas Law:

• \(P = 150\) atm
• \(V = 10.0\) L
• \(R = 0.0821\) L·atm·mol⁻¹·K⁻¹
• \(T = 293\) K

[
PV = nRT \
(150)(10.0) = n(0.0821)(293) \
1500 = n(24.06) \
n = \frac{1500}{24.06} \approx 62.4\,\text{mol}
]

Step 2: Convert patient flow rate to moles per minute

At 1.0 atm and 20 °C, the gas the patient breathes also obeys the Ideal Gas Law.

For 2.0 L/min:

[
PV = nRT \
(1.0)(2.0) = n(0.0821)(293) \
2.0 = n(24.06) \
n = \frac{2.0}{24.06} \approx 0.0831\,\text{mol/min}
]

Step 3: Time until the cylinder is empty

[
\text{time} = \frac{62.4\,\text{mol}}{0.0831\,\text{mol/min}} \approx 751\,\text{min}
]

That’s about 12.5 hours of continuous use at that flow rate. This is one of the more realistic examples of gas law problems with solutions you might see in nursing, respiratory therapy, or pre-med courses.

Example 3C: Estimating moles of air in a room

Let’s scale up.

Problem
Estimate how many moles of air are in a small classroom that measures 30 ft × 20 ft × 10 ft at 77 °F and 1.0 atm.

Step 1: Convert volume to liters

First find cubic feet:

[
V = 30 \times 20 \times 10 = 6000\,\text{ft}^3
]

Use 1 ft³ ≈ 28.3 L:

[
V \approx 6000 \times 28.3 = 169{,}800\,\text{L}
]

Step 2: Convert temperature

77 °F → 25 °C → 298 K.

Step 3: Apply Ideal Gas Law

• \(P = 1.0\) atm
• \(V = 1.698 \times 10^5\) L
• \(T = 298\) K
• \(R = 0.0821\)

[
PV = nRT \
(1.0)(1.698 \times 10^5) = n(0.0821)(298) \
1.698 \times 10^5 = n(24.48) \
n \approx \frac{1.698 \times 10^5}{24.48} \approx 6935\,\text{mol}
]

That’s about 7.0 × 10³ moles of air. This order-of-magnitude estimate is handy in environmental science and ventilation design.

Pulling it together: why these 3 practical examples matter

When teachers assign examples of gas law problems with solutions: 3 practical examples, they’re not just being traditional. These three core situations—compression (Boyle), heating/cooling (Charles), and full-system changes (Ideal Gas Law)—cover most of what you’ll see in chemistry, physics, and even engineering.

Some patterns to watch for in future problems:

• If pressure and volume change but temperature is constant, think Boyle’s Law.
• If volume and temperature change at constant pressure, think Charles’s Law.
• If the problem mentions moles or asks “how many grams,” you’re probably in Ideal Gas Law territory.
• Many modern real examples, from ventilators to hydrogen fuel tanks, still boil down to these same relationships.

Modern data on air quality, climate, and health—like the EPA’s air trends (https://www.epa.gov/air-trends) or NIH respiratory research—often rely on the same gas law reasoning behind these classroom problems. That’s why working through several examples of gas law problems with solutions is more than just exam prep; it’s training for how scientists and engineers think about the air around us.

FAQ: Quick answers about examples of gas law problems

Q1. What are some common examples of gas law problems with solutions used in high school chemistry?
Common classroom examples of gas law problems include compressing air in a syringe (Boyle’s Law), heating a gas in a sealed container (pressure–temperature relationship), a hot-air balloon rising as its air warms (Charles’s Law), and using the Ideal Gas Law to find moles of gas in a scuba tank or oxygen cylinder.

Q2. How do I know which gas law to use in a given example of a gas law problem?
Look at what changes and what stays constant. If temperature is constant and pressure/volume change, use Boyle’s Law. If pressure is constant and volume/temperature change, use Charles’s Law. If the problem involves moles, mass, or needs a more complete description, use the Ideal Gas Law. Many of the best examples of gas law problems with solutions: 3 practical examples are designed to train this pattern recognition.

Q3. Are there real examples of gas law applications in medicine or health?
Yes. Oxygen cylinders for patients, anesthetic gas delivery in operating rooms, and ventilator settings in intensive care all rely on gas laws. Organizations like the Mayo Clinic (https://www.mayoclinic.org) and NIH fund research that depends on accurate modeling of gas volumes, pressures, and flows.

Q4. Do real gases always follow these examples perfectly?
Not perfectly. At very high pressures or very low temperatures, real gases deviate from ideal behavior. But for most classroom examples of gas law problems with solutions, the ideal gas approximation is accurate enough to teach the core ideas and give reasonable numerical answers.

Q5. Where can I find more practice problems beyond these 3 practical examples?
Many university chemistry departments host free practice sets. For instance, general chemistry courses at major universities (such as those linked from https://ocw.mit.edu or large state schools’ .edu sites) often include additional examples of gas law problems with solutions you can download as PDFs.