Combined Gas Law Word Problems Worksheet
One-page worksheet with six combined gas law word problems for chemistry students, featuring a light blue background and a "Keystone Science" logo.
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Step-by-step solution for: Chemistry Combined Gas Law Problems with Answer Key - Keystone Science
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Step-by-step solution for: Chemistry Combined Gas Law Problems with Answer Key - Keystone Science
Here are the step-by-step solutions for the Combined Gas Law problems.
The formula we use is:
$$ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $$
Important Rule: Temperature must always be in Kelvin ($K$). To convert Celsius to Kelvin, add 273 ($^\circ C + 273 = K$).
---
* Given:
* $P_1 = 120 \text{ kPa}$
* $V_1 = 45 \text{ L}$
* $T_1 = 81^\circ\text{C} \rightarrow 81 + 273 = 354 \text{ K}$
* $P_2 = 50 \text{ kPa}$
* $V_2 = 40 \text{ L}$
* $T_2 = ?$
* Setup:
$$ \frac{120 \times 45}{354} = \frac{50 \times 40}{T_2} $$
* Solve:
1. Left side: $(120 \times 45) / 354 = 5400 / 354 \approx 15.254$
2. Right side numerator: $50 \times 40 = 2000$
3. Equation: $15.254 = 2000 / T_2$
4. Rearrange for $T_2$: $T_2 = 2000 / 15.254 \approx 131.1 \text{ K}$
5. Convert back to Celsius: $131.1 - 273 = -141.9^\circ\text{C}$
Answer: $-142^\circ\text{C}$ (rounded)
---
* Given:
* $V_1 = 21,000 \text{ L}$
* $P_1 = 100 \text{ kPa}$
* $V_2 = 14,000 \text{ L}$
* $P_2 = 150 \text{ kPa}$
* $T_2 = 300 \text{ K}$
* $T_1 = ?$
* Setup:
$$ \frac{100 \times 21,000}{T_1} = \frac{150 \times 14,000}{300} $$
* Solve:
1. Right side: $(150 \times 14,000) / 300 = 2,100,000 / 300 = 7,000$
2. Left side numerator: $100 \times 21,000 = 2,100,000$
3. Equation: $2,100,000 / T_1 = 7,000$
4. Rearrange for $T_1$: $T_1 = 2,100,000 / 7,000 = 300 \text{ K}$
Answer: $300 \text{ K}$
---
* Given:
* $T_1 = 500 \text{ K}$
* $P_1 = 280 \text{ kPa}$
* $V_1 = 18 \text{ L}$
* $T_2 = 350 \text{ K}$
* $P_2 = 380 \text{ kPa}$
* $V_2 = ?$
* Setup:
$$ \frac{280 \times 18}{500} = \frac{380 \times V_2}{350} $$
* Solve:
1. Left side: $(280 \times 18) / 500 = 5,040 / 500 = 10.08$
2. Equation: $10.08 = (380 \times V_2) / 350$
3. Multiply both sides by 350: $3,528 = 380 \times V_2$
4. Divide by 380: $V_2 = 3,528 / 380 \approx 9.28 \text{ L}$
Answer: $9.28 \text{ L}$
---
* Given:
* $P_1 = 75 \text{ kPa}$
* $T_1 = 27^\circ\text{C} \rightarrow 27 + 273 = 300 \text{ K}$
* $V_1 = 12 \text{ L}$
* $P_2 = 53 \text{ kPa}$
* $T_2 = 93^\circ\text{C} \rightarrow 93 + 273 = 366 \text{ K}$
* $V_2 = ?$
* Setup:
$$ \frac{75 \times 12}{300} = \frac{53 \times V_2}{366} $$
* Solve:
1. Left side: $(75 \times 12) / 300 = 900 / 300 = 3$
2. Equation: $3 = (53 \times V_2) / 366$
3. Multiply both sides by 366: $1,098 = 53 \times V_2$
4. Divide by 53: $V_2 = 1,098 / 53 \approx 20.72 \text{ L}$
Answer: $20.7 \text{ L}$
---
* Given:
* $V_1 = 5 \text{ L}$
* $T_1 = 2000 \text{ K}$
* $P_1 = 1500 \text{ mmHg}$
* $V_2 = 25 \text{ L}$
* $T_2 = 1750 \text{ K}$
* $P_2 = ?$
* Setup:
$$ \frac{1500 \times 5}{2000} = \frac{P_2 \times 25}{1750} $$
* Solve:
1. Left side: $(1500 \times 5) / 2000 = 7,500 / 2,000 = 3.75$
2. Equation: $3.75 = (P_2 \times 25) / 1750$
3. Multiply both sides by 1750: $6,562.5 = P_2 \times 25$
4. Divide by 25: $P_2 = 6,562.5 / 25 = 262.5 \text{ mmHg}$
Answer: $262.5 \text{ mmHg}$
---
* Given:
* $T_1 = 212^\circ\text{C} \rightarrow 212 + 273 = 485 \text{ K}$
* $V_1 = 30 \text{ L}$
* $T_2 = 380^\circ\text{C} \rightarrow 380 + 273 = 653 \text{ K}$
* $V_2 = 18 \text{ L}$
* $P_2 = 1.85 \text{ atm}$
* $P_1 = ?$
* Setup:
$$ \frac{P_1 \times 30}{485} = \frac{1.85 \times 18}{653} $$
* Solve:
1. Right side: $(1.85 \times 18) / 653 = 33.3 / 653 \approx 0.051$
2. Equation: $(P_1 \times 30) / 485 = 0.051$
3. Multiply both sides by 485: $P_1 \times 30 = 24.735$
4. Divide by 30: $P_1 = 24.735 / 30 \approx 0.82 \text{ atm}$
Answer: $0.82 \text{ atm}$
Final Answer:
1. -142°C
2. 300 K
3. 9.28 L
4. 20.7 L
5. 262.5 mmHg
6. 0.82 atm
The formula we use is:
$$ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $$
Important Rule: Temperature must always be in Kelvin ($K$). To convert Celsius to Kelvin, add 273 ($^\circ C + 273 = K$).
---
Problem 1
* Given:
* $P_1 = 120 \text{ kPa}$
* $V_1 = 45 \text{ L}$
* $T_1 = 81^\circ\text{C} \rightarrow 81 + 273 = 354 \text{ K}$
* $P_2 = 50 \text{ kPa}$
* $V_2 = 40 \text{ L}$
* $T_2 = ?$
* Setup:
$$ \frac{120 \times 45}{354} = \frac{50 \times 40}{T_2} $$
* Solve:
1. Left side: $(120 \times 45) / 354 = 5400 / 354 \approx 15.254$
2. Right side numerator: $50 \times 40 = 2000$
3. Equation: $15.254 = 2000 / T_2$
4. Rearrange for $T_2$: $T_2 = 2000 / 15.254 \approx 131.1 \text{ K}$
5. Convert back to Celsius: $131.1 - 273 = -141.9^\circ\text{C}$
Answer: $-142^\circ\text{C}$ (rounded)
---
Problem 2
* Given:
* $V_1 = 21,000 \text{ L}$
* $P_1 = 100 \text{ kPa}$
* $V_2 = 14,000 \text{ L}$
* $P_2 = 150 \text{ kPa}$
* $T_2 = 300 \text{ K}$
* $T_1 = ?$
* Setup:
$$ \frac{100 \times 21,000}{T_1} = \frac{150 \times 14,000}{300} $$
* Solve:
1. Right side: $(150 \times 14,000) / 300 = 2,100,000 / 300 = 7,000$
2. Left side numerator: $100 \times 21,000 = 2,100,000$
3. Equation: $2,100,000 / T_1 = 7,000$
4. Rearrange for $T_1$: $T_1 = 2,100,000 / 7,000 = 300 \text{ K}$
Answer: $300 \text{ K}$
---
Problem 3
* Given:
* $T_1 = 500 \text{ K}$
* $P_1 = 280 \text{ kPa}$
* $V_1 = 18 \text{ L}$
* $T_2 = 350 \text{ K}$
* $P_2 = 380 \text{ kPa}$
* $V_2 = ?$
* Setup:
$$ \frac{280 \times 18}{500} = \frac{380 \times V_2}{350} $$
* Solve:
1. Left side: $(280 \times 18) / 500 = 5,040 / 500 = 10.08$
2. Equation: $10.08 = (380 \times V_2) / 350$
3. Multiply both sides by 350: $3,528 = 380 \times V_2$
4. Divide by 380: $V_2 = 3,528 / 380 \approx 9.28 \text{ L}$
Answer: $9.28 \text{ L}$
---
Problem 4
* Given:
* $P_1 = 75 \text{ kPa}$
* $T_1 = 27^\circ\text{C} \rightarrow 27 + 273 = 300 \text{ K}$
* $V_1 = 12 \text{ L}$
* $P_2 = 53 \text{ kPa}$
* $T_2 = 93^\circ\text{C} \rightarrow 93 + 273 = 366 \text{ K}$
* $V_2 = ?$
* Setup:
$$ \frac{75 \times 12}{300} = \frac{53 \times V_2}{366} $$
* Solve:
1. Left side: $(75 \times 12) / 300 = 900 / 300 = 3$
2. Equation: $3 = (53 \times V_2) / 366$
3. Multiply both sides by 366: $1,098 = 53 \times V_2$
4. Divide by 53: $V_2 = 1,098 / 53 \approx 20.72 \text{ L}$
Answer: $20.7 \text{ L}$
---
Problem 5
* Given:
* $V_1 = 5 \text{ L}$
* $T_1 = 2000 \text{ K}$
* $P_1 = 1500 \text{ mmHg}$
* $V_2 = 25 \text{ L}$
* $T_2 = 1750 \text{ K}$
* $P_2 = ?$
* Setup:
$$ \frac{1500 \times 5}{2000} = \frac{P_2 \times 25}{1750} $$
* Solve:
1. Left side: $(1500 \times 5) / 2000 = 7,500 / 2,000 = 3.75$
2. Equation: $3.75 = (P_2 \times 25) / 1750$
3. Multiply both sides by 1750: $6,562.5 = P_2 \times 25$
4. Divide by 25: $P_2 = 6,562.5 / 25 = 262.5 \text{ mmHg}$
Answer: $262.5 \text{ mmHg}$
---
Problem 6
* Given:
* $T_1 = 212^\circ\text{C} \rightarrow 212 + 273 = 485 \text{ K}$
* $V_1 = 30 \text{ L}$
* $T_2 = 380^\circ\text{C} \rightarrow 380 + 273 = 653 \text{ K}$
* $V_2 = 18 \text{ L}$
* $P_2 = 1.85 \text{ atm}$
* $P_1 = ?$
* Setup:
$$ \frac{P_1 \times 30}{485} = \frac{1.85 \times 18}{653} $$
* Solve:
1. Right side: $(1.85 \times 18) / 653 = 33.3 / 653 \approx 0.051$
2. Equation: $(P_1 \times 30) / 485 = 0.051$
3. Multiply both sides by 485: $P_1 \times 30 = 24.735$
4. Divide by 30: $P_1 = 24.735 / 30 \approx 0.82 \text{ atm}$
Answer: $0.82 \text{ atm}$
Final Answer:
1. -142°C
2. 300 K
3. 9.28 L
4. 20.7 L
5. 262.5 mmHg
6. 0.82 atm
Parent Tip: Review the logic above to help your child master the concept of combined gas law worksheet.
