Gas Laws Worksheet Generator
Boyle's, Charles's, Gay-Lussac's, combined, ideal, Dalton's, and molar volume. No AI, no credits.
Grades 9 – 12No account needed
Problem sets for Boyle's, Charles's, Gay-Lussac's, the combined gas law, the ideal gas law, Dalton's law of partial pressures, and molar volume. Constants (including R in each unit set) come from stored NIST data. The assumptions block states the standard conditions and the value of R.
What students practice
- Chemistry
- Gas laws
- Ideal gas law
Live sample sheet
No account needed to view and print.
1. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
2. A gas occupies 5 L at 300.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 300.15 K · V2 = 5 L
3. How many liters does 3 mol of gas occupy at 3 atm and 348 K?
n = 3 mol · P = 3 atm · T = 348 K · R = 0.082057 L·atm/(mol·K)
4. A gas occupies 3 L at 2 atm. What is the volume at 3 atm (constant temperature)?
P1 = 2 atm · V1 = 3 L · P2 = 3 atm
5. A gas occupies 5 L at 326.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 326.15 K · V2 = 5 L
6. How many liters does 1 mol of gas occupy at 1 atm and 310 K?
n = 1 mol · P = 1 atm · T = 310 K · R = 0.082057 L·atm/(mol·K)
7. A gas occupies 2 L at 1 atm. What is the volume at 2 atm (constant temperature)?
P1 = 1 atm · V1 = 2 L · P2 = 2 atm
8. A gas occupies 2 L at 329.15 K. What is the temperature at 2 L (constant pressure)?
V1 = 2 L · T1 = 329.15 K · V2 = 2 L
9. How many liters does 1 mol of gas occupy at 2 atm and 335 K?
n = 1 mol · P = 2 atm · T = 335 K · R = 0.082057 L·atm/(mol·K)
10. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
1. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 5 = 2 × V2
V2 = (2 × 5) / 2 = 5 L
Answer: 5 L
Verified: direct formula vs algebraic rearrangement
2. A gas occupies 5 L at 300.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 300.15 K · V2 = 5 L
Solution:
Charles's law: V1/T1 = V2/T2
5/300.15 = 5/T2
T2 = (5 × 300.15) / 5 = 300.15 K
Answer: 300.15 K
Verified: direct formula vs algebraic rearrangement
3. How many liters does 3 mol of gas occupy at 3 atm and 348 K?
n = 3 mol · P = 3 atm · T = 348 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (3 × 0.082057 × 348) / 3 = 28.56 L
Answer: 28.56 L
Verified: direct formula vs algebraic rearrangement
4. A gas occupies 3 L at 2 atm. What is the volume at 3 atm (constant temperature)?
P1 = 2 atm · V1 = 3 L · P2 = 3 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 3 = 3 × V2
V2 = (2 × 3) / 3 = 2 L
Answer: 2 L
Verified: direct formula vs algebraic rearrangement
5. A gas occupies 5 L at 326.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 326.15 K · V2 = 5 L
Solution:
Charles's law: V1/T1 = V2/T2
5/326.15 = 5/T2
T2 = (5 × 326.15) / 5 = 326.15 K
Answer: 326.15 K
Verified: direct formula vs algebraic rearrangement
6. How many liters does 1 mol of gas occupy at 1 atm and 310 K?
n = 1 mol · P = 1 atm · T = 310 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (1 × 0.082057 × 310) / 1 = 25.44 L
Answer: 25.44 L
Verified: direct formula vs algebraic rearrangement
7. A gas occupies 2 L at 1 atm. What is the volume at 2 atm (constant temperature)?
P1 = 1 atm · V1 = 2 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
1 × 2 = 2 × V2
V2 = (1 × 2) / 2 = 1 L
Answer: 1 L
Verified: direct formula vs algebraic rearrangement
8. A gas occupies 2 L at 329.15 K. What is the temperature at 2 L (constant pressure)?
V1 = 2 L · T1 = 329.15 K · V2 = 2 L
Solution:
Charles's law: V1/T1 = V2/T2
2/329.15 = 2/T2
T2 = (2 × 329.15) / 2 = 329.15 K
Answer: 329.15 K
Verified: direct formula vs algebraic rearrangement
9. How many liters does 1 mol of gas occupy at 2 atm and 335 K?
n = 1 mol · P = 2 atm · T = 335 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (1 × 0.082057 × 335) / 2 = 13.74 L
Answer: 13.74 L
Verified: direct formula vs algebraic rearrangement
10. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 5 = 2 × V2
V2 = (2 × 5) / 2 = 5 L
Answer: 5 L
Verified: direct formula vs algebraic rearrangement
Assumption: R = 0.082057 L·atm/(mol·K) (NIST).
Assumption: STP: 0°C (273.15 K) and 1 atm.
Assumption: Molar volume at STP = 22.414 L/mol.
Convention: Pressure in atm, volume in L, temperature in K.
Convention: Round answers to 2 decimal places.
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Frequently asked questions
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Every answer is computed by code and verified by a second independent method. No AI is used.
What grades is the Gas Laws for?
This tool fits Grades 9 – 12.
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1. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
2. A gas occupies 5 L at 300.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 300.15 K · V2 = 5 L
3. How many liters does 3 mol of gas occupy at 3 atm and 348 K?
n = 3 mol · P = 3 atm · T = 348 K · R = 0.082057 L·atm/(mol·K)
4. A gas occupies 3 L at 2 atm. What is the volume at 3 atm (constant temperature)?
P1 = 2 atm · V1 = 3 L · P2 = 3 atm
5. A gas occupies 5 L at 326.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 326.15 K · V2 = 5 L
6. How many liters does 1 mol of gas occupy at 1 atm and 310 K?
n = 1 mol · P = 1 atm · T = 310 K · R = 0.082057 L·atm/(mol·K)
7. A gas occupies 2 L at 1 atm. What is the volume at 2 atm (constant temperature)?
P1 = 1 atm · V1 = 2 L · P2 = 2 atm
8. A gas occupies 2 L at 329.15 K. What is the temperature at 2 L (constant pressure)?
V1 = 2 L · T1 = 329.15 K · V2 = 2 L
9. How many liters does 1 mol of gas occupy at 2 atm and 335 K?
n = 1 mol · P = 2 atm · T = 335 K · R = 0.082057 L·atm/(mol·K)
10. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
1. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 5 = 2 × V2
V2 = (2 × 5) / 2 = 5 L
Answer: 5 L
Verified: direct formula vs algebraic rearrangement
2. A gas occupies 5 L at 300.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 300.15 K · V2 = 5 L
Solution:
Charles's law: V1/T1 = V2/T2
5/300.15 = 5/T2
T2 = (5 × 300.15) / 5 = 300.15 K
Answer: 300.15 K
Verified: direct formula vs algebraic rearrangement
3. How many liters does 3 mol of gas occupy at 3 atm and 348 K?
n = 3 mol · P = 3 atm · T = 348 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (3 × 0.082057 × 348) / 3 = 28.56 L
Answer: 28.56 L
Verified: direct formula vs algebraic rearrangement
4. A gas occupies 3 L at 2 atm. What is the volume at 3 atm (constant temperature)?
P1 = 2 atm · V1 = 3 L · P2 = 3 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 3 = 3 × V2
V2 = (2 × 3) / 3 = 2 L
Answer: 2 L
Verified: direct formula vs algebraic rearrangement
5. A gas occupies 5 L at 326.15 K. What is the temperature at 5 L (constant pressure)?
V1 = 5 L · T1 = 326.15 K · V2 = 5 L
Solution:
Charles's law: V1/T1 = V2/T2
5/326.15 = 5/T2
T2 = (5 × 326.15) / 5 = 326.15 K
Answer: 326.15 K
Verified: direct formula vs algebraic rearrangement
6. How many liters does 1 mol of gas occupy at 1 atm and 310 K?
n = 1 mol · P = 1 atm · T = 310 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (1 × 0.082057 × 310) / 1 = 25.44 L
Answer: 25.44 L
Verified: direct formula vs algebraic rearrangement
7. A gas occupies 2 L at 1 atm. What is the volume at 2 atm (constant temperature)?
P1 = 1 atm · V1 = 2 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
1 × 2 = 2 × V2
V2 = (1 × 2) / 2 = 1 L
Answer: 1 L
Verified: direct formula vs algebraic rearrangement
8. A gas occupies 2 L at 329.15 K. What is the temperature at 2 L (constant pressure)?
V1 = 2 L · T1 = 329.15 K · V2 = 2 L
Solution:
Charles's law: V1/T1 = V2/T2
2/329.15 = 2/T2
T2 = (2 × 329.15) / 2 = 329.15 K
Answer: 329.15 K
Verified: direct formula vs algebraic rearrangement
9. How many liters does 1 mol of gas occupy at 2 atm and 335 K?
n = 1 mol · P = 2 atm · T = 335 K · R = 0.082057 L·atm/(mol·K)
Solution:
Ideal gas law: PV = nRT
V = nRT/P
V = (1 × 0.082057 × 335) / 2 = 13.74 L
Answer: 13.74 L
Verified: direct formula vs algebraic rearrangement
10. A gas occupies 5 L at 2 atm. What is the volume at 2 atm (constant temperature)?
P1 = 2 atm · V1 = 5 L · P2 = 2 atm
Solution:
Boyle's law: P1V1 = P2V2
2 × 5 = 2 × V2
V2 = (2 × 5) / 2 = 5 L
Answer: 5 L
Verified: direct formula vs algebraic rearrangement
Assumption: R = 0.082057 L·atm/(mol·K) (NIST).
Assumption: STP: 0°C (273.15 K) and 1 atm.
Assumption: Molar volume at STP = 22.414 L/mol.
Convention: Pressure in atm, volume in L, temperature in K.
Convention: Round answers to 2 decimal places.
Gas Laws - Chemistry - Mixed gas laws
Boyle's, Charles's, Gay-Lussac's, combined, ideal, Dalton's, and molar volume. No AI, no credits.