Ideal Gas Law worksheet with practice problems and formulas.

Let’s solve each problem one by one using the Ideal Gas Law:
PV = nRT

Where:
- P = pressure (in atm)
- V = volume (in liters)
- n = number of moles
- R = gas constant = 0.0821 L·atm/mol·K
- T = temperature in Kelvin (°C + 273)

We’ll rearrange the formula as needed for each question.

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Problem 1:
How many moles of nitrogen gas occupy a volume of 42.0 liters at 2.00 atm and 25°C?

Step 1: Convert °C to K → 25 + 273 = 298 K
Step 2: Use PV = nRT → solve for n → n = PV / RT
Plug in:
P = 2.00 atm
V = 42.0 L
R = 0.0821
T = 298 K

n = (2.00 × 42.0) / (0.0821 × 298)
= 84.0 / 24.4658 ≈ 3.43 moles

✔ Answer: 3.43 mol

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Problem 2:
What volume will 2.00 moles of oxygen gas occupy at 720 torr and 25°C?

Note: Pressure is in torr — convert to atm!
760 torr = 1 atm → 720 torr = 720/760 ≈ 0.9474 atm

T = 25 + 273 = 298 K

Use PV = nRT → solve for V → V = nRT / P

n = 2.00 mol
R = 0.0821
T = 298 K
P = 0.9474 atm

V = (2.00 × 0.0821 × 298) / 0.9474
= (48.9316) / 0.9474 ≈ 51.65 L

✔ Answer: 51.7 L (rounded to 3 sig figs)

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Problem 3:
What pressure is exerted by 25.0 g of CO₂ at 25°C in a container with a volume of 500 mL?

First, find moles of CO₂:
Molar mass CO₂ = 12 + 16×2 = 44 g/mol
n = 25.0 g / 44 g/mol ≈ 0.5682 mol

Volume = 500 mL = 0.500 L
T = 25 + 273 = 298 K

Use PV = nRT → solve for P → P = nRT / V

P = (0.5682 × 0.0821 × 298) / 0.500
= (13.90) / 0.500 ≈ 27.8 atm

✔ Answer: 27.8 atm

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Problem 4:
At what temperature will 0.500 mol of Cl₂ exert a pressure of 400 torr in a container of 2.00 L?

Convert pressure: 400 torr = 400/760 ≈ 0.5263 atm

Use PV = nRT → solve for T → T = PV / nR

P = 0.5263 atm
V = 2.00 L
n = 0.500 mol
R = 0.0821

T = (0.5263 × 2.00) / (0.500 × 0.0821)
= 1.0526 / 0.04105 ≈ 25.64 K

Wait — that’s way too cold! Let me double-check:

Actually, 0.5263 × 2.00 = 1.0526
0.500 × 0.0821 = 0.04105
1.0526 ÷ 0.04105 ≈ 25.64 K? That can’t be right — chlorine would be solid!

I think I made a mistake in units or calculation.

Wait — let’s recalculate carefully:

T = (P × V) / (n × R)
= (0.5263 atm × 2.00 L) / (0.500 mol × 0.0821 L·atm/mol·K)
= 1.0526 / 0.04105 ≈ 25.64 K

That’s mathematically correct but physically unrealistic. Maybe the problem meant 4000 torr? Or perhaps it's a trick? But let’s assume the numbers are correct.

But wait — maybe I misread the problem. Let me check again.

“0.500 mol of Cl₂ exert a pressure of 400 torr in a container of 2.00 L”

Yes — so unless it’s supercooled, this doesn’t make sense. But since we’re doing math, we’ll go with it.

So T ≈ 25.6 K

Convert to °C: 25.6 - 273 = -247.4°C

But the question asks for temperature — probably expects Kelvin.

✔ Answer: 25.6 K

*(Note: This is extremely cold — below boiling point of liquid nitrogen — so likely an error in problem setup, but mathematically correct based on given values.)*

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Problem 5:
What is the molarity of NH₃ if 200 mL are at 25°C and 1.00 atm?

Molarity = moles per liter → first find moles, then divide by volume in L.

V = 200 mL = 0.200 L
T = 298 K
P = 1.00 atm

Use PV = nRT → n = PV / RT
n = (1.00 × 0.200) / (0.0821 × 298)
= 0.200 / 24.4658 ≈ 0.008175 mol

Now, molarity = moles / liters = 0.008175 mol / 0.200 L = 0.0409 M

✔ Answer: 0.0409 M

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Problem 6:
What is the density of O₂ gas at STP?

STP = Standard Temperature and Pressure = 0°C (273 K) and 1 atm

Density = mass / volume

For gases, we can use:
d = (PM) / (RT) where M = molar mass

O₂ molar mass = 32.0 g/mol
P = 1 atm
R = 0.0821
T = 273 K

d = (1 × 32.0) / (0.0821 × 273)
= 32.0 / 22.4133 ≈ 1.428 g/L

✔ Answer: 1.43 g/L (rounded)

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Problem 7:
What volume of N₂ gas at 25°C and 750 torr contains 0.500 mole?

T = 298 K
P = 750 torr = 750/760 ≈ 0.9868 atm
n = 0.500 mol

V = nRT / P
= (0.500 × 0.0821 × 298) / 0.9868
= (12.2329) / 0.9868 ≈ 12.40 L

✔ Answer: 12.4 L

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Problem 8:
What volume will 20.0 grams of H₂ hydrogen gas occupy at 25°C and 1.00 atm?

Molar mass H₂ = 2.016 g/mol ≈ 2.02 g/mol
n = 20.0 g / 2.02 g/mol ≈ 9.901 mol

T = 298 K
P = 1.00 atm

V = nRT / P
= (9.901 × 0.0821 × 298) / 1.00
= (242.2) / 1.00 ≈ 242 L

✔ Answer: 242 L

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Problem 9:
Find the number of grams of C₂H₂ that exert a pressure of 780 torr in a volume of 500 mL at a temperature of 25°C.

First, convert:
P = 780 torr = 780/760 ≈ 1.0263 atm
V = 500 mL = 0.500 L
T = 298 K

Use PV = nRT → n = PV / RT
n = (1.0263 × 0.500) / (0.0821 × 298)
= 0.51315 / 24.4658 ≈ 0.02097 mol

Molar mass C₂H₂ = 2×12 + 2×1 = 26 g/mol
Mass = moles × molar mass = 0.02097 × 26 ≈ 0.545 g

✔ Answer: 0.545 g

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Problem 10:
The temperature of a gas increases from 25°C to 50°C, while its volume remains constant. If the initial pressure was 1.00 atm, what is the final pressure?

Since volume and moles are constant, use Gay-Lussac’s Law:
P₁/T₁ = P₂/T₂ → P₂ = P₁ × (T₂/T₁)

T₁ = 25 + 273 = 298 K
T₂ = 50 + 273 = 323 K
P₁ = 1.00 atm

P₂ = 1.00 × (323 / 298) ≈ 1.00 × 1.0839 ≈ 1.08 atm

✔ Answer: 1.08 atm

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Final Answer:
1. 3.43 mol
2. 51.7 L
3. 27.8 atm
4. 25.6 K
5. 0.0409 M
6. 1.43 g/L
7. 12.4 L
8. 242 L
9. 0.545 g
10. 1.08 atm