MCAT Chemistry (Grade 11 Chemistry) Style Questions
Question 1: What will be the pressure of the gas at 200 K if the volume does not change?
Given:
- Initial pressure: \(P_1 = 80.0 , \text{mmHg}\)
- Initial temperature: \(T_1 = 700 , \text{K}\)
- Final temperature: \(T_2 = 200 , \text{K}\)
- Final pressure: \(P_2 = ?\)
Solution:
This is an application of Gay-Lussac’s Law, which states:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)Rearranging to solve for \(P_2\):
\(P_2 = P_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(P_2 = 80.0 \cdot \frac{200}{700}\)
\(P_2 = 80.0 \cdot 0.2857\)
\(P_2 = 22.9 , \text{mmHg}\)
Answer: The final pressure is 22.9 mmHg.
Question 2: What is the temperature if the pressure of argon gas rises to 750 mmHg?
Given:
- Initial pressure: \(P_1 = 100.0 , \text{mmHg}\)
- Initial temperature: \(T_1 = 327^\circ \text{C} = 600 , \text{K}\)
- Final pressure: \(P_2 = 750 , \text{mmHg}\)
- Final temperature: \(T_2 = ?\)
Solution:
Using Gay-Lussac’s Law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)Rearranging to solve for \(T_2\):
\(T_2 = T_1 \cdot \frac{P_2}{P_1}\)Substitute the given values:
\(T_2 = 600 \cdot \frac{750}{100.0}\)
\(T_2 = 600 \cdot 7.5\)
\(T_2 = 4500 , \text{K}\)
To convert to Celsius:
\(T_2 = 4500 – 273.15\)
\(T_2 = 4226.85^\circ \text{C}\)
Answer: The final temperature is 4500 K or 4227°C.
Question 3: What is the new pressure in the car tire?
Given:
- Initial pressure: \(P_1 = 2.0 , \text{atm}\)
- Initial temperature: \(T_1 = 27^\circ\text{C} = 300 , \text{K}\)
- Final temperature: \(T_2 = 35^\circ\text{C} = 308 , \text{K}\)
- Final pressure: \(P_2 = ?\)
Solution:
Using Gay-Lussac’s Law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)Rearranging to solve for \(P_2\):
\(P_2 = P_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(P_2 = 2.0 \cdot \frac{308}{300}\)
\(P_2 = 2.0 \cdot 1.0267\)
\(P_2 = 2.1 , \text{atm}\)
Answer: The new pressure in the tire is 2.1 atm.
Question 4: Will the glass Christmas ornament burst?
Given:
- Initial pressure: \(P_1 = 1.01 , \text{atm}\)
- Initial temperature: \(T_1 = 25^\circ\text{C} = 298.15 , \text{K}\)
- Final temperature: \(T_2 = 250^\circ\text{C} = 523.15 , \text{K}\)
- Final pressure: \(P_2 = ?\)
- Maximum pressure before bursting: \(P_{\text{max}} = 4.30 , \text{atm}\)
Solution:
Using Gay-Lussac’s Law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)Rearranging to solve for \(P_2\):
\(P_2 = P_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(P_2 = 1.01 \cdot \frac{523.15}{298.15}\)
\(P_2 = 1.01 \cdot 1.7535\)
\(P_2 = 1.77 , \text{atm}\)
Since \(P_2 = 1.77 , \text{atm}\) is less than \(P_{\text{max}} = 4.30 , \text{atm}\), the ornament will not burst.
Answer: The final pressure is 1.77 atm, so the ornament will not burst.
Question 5: What will be the pressure in the spray can when heated?
Given:
- Initial pressure: \(P_1 = 1.2 , \text{atm}\)
- Initial temperature: \(T_1 = 24^\circ\text{C} = 297.15 , \text{K}\)
- Final temperature: \(T_2 = 485^\circ\text{C} = 758.15 , \text{K}\)
- Final pressure: \(P_2 = ?\)
Solution:
Using Gay-Lussac’s Law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)Rearranging to solve for \(P_2\):
\(P_2 = P_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(P_2 = 1.2 \cdot \frac{758.15}{297.15}\)
\(P_2 = 1.2 \cdot 2.551\)
\(P_2 = 3.1 , \text{atm}\)
Answer: The pressure inside the can will be 3.1 atm.
Here are detailed step-by-step solutions for each question:
Question 6: What will be the volume of argon gas at 27°C if the pressure does not change?
Given:
- Initial volume: \(V_1 = 7.5 , \text{L}\)
- Initial temperature: \(T_1 = 327^\circ \text{C} = 600 , \text{K}\)
- Final temperature: \(T_2 = 27^\circ \text{C} = 300 , \text{K}\)
- Final volume: \(V_2 = ?\)
Solution:
This is an application of Charles’s Law, which states:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(V_2\):
\(V_2 = V_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(V_2 = 7.5 \cdot \frac{300}{600}\)
\(V_2 = 7.5 \cdot 0.5\)
\(V_2 = 3.8 , \text{L}\)
Answer: The final volume is 3.8 L.
Question 7: At what temperature will the volume of helium gas increase to 15 L?
Given:
- Initial volume: \(V_1 = 5.0 , \text{L}\)
- Initial temperature: \(T_1 = 298 , \text{K}\)
- Final volume: \(V_2 = 15 , \text{L}\)
- Final temperature: \(T_2 = ?\)
Solution:
Using Charles’s Law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(T_2\):
\(T_2 = T_1 \cdot \frac{V_2}{V_1}\)Substitute the given values:
\(T_2 = 298 \cdot \frac{15}{5.0}\)
\(T_2 = 298 \cdot 3\)
\(T_2 = 894 , \text{K}\)
Convert to Celsius:
\(T_2 = 894 – 273.15\)
\(T_2 = 620.85^\circ \text{C}\)
Answer: The final temperature is approximately 621°C.
Question 8: What is the new volume of air warmed from -50°C to 120°C?
Given:
- Initial volume: \(V_1 = 12 , \text{L}\)
- Initial temperature: \(T_1 = -50^\circ \text{C} = 223.15 , \text{K}\)
- Final temperature: \(T_2 = 120^\circ \text{C} = 393.15 , \text{K}\)
- Final volume: \(V_2 = ?\)
Solution:
Using Charles’s Law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(V_2\):
\(V_2 = V_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(V_2 = 12 \cdot \frac{393.15}{223.15}\)
\(V_2 = 12 \cdot 1.762\)
\(V_2 = 21.1 , \text{L}\)
Answer: The final volume is approximately 21 L.
Question 9: At what temperature will a gas have half the volume of its room temperature volume?
Given:
- Initial temperature: \(T_1 = 24.0^\circ \text{C} = 297.15 , \text{K}\)
- Final volume: \(V_2 = 0.5 , V_1\)
- Final temperature: \(T_2 = ?\)
Solution:
Using Charles’s Law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(T_2\):
\(T_2 = T_1 \cdot \frac{V_2}{V_1}\)Substitute the given values:
\(T_2 = 297.15 \cdot \frac{0.5}{1}\)
\(T_2 = 297.15 \cdot 0.5\)
\(T_2 = 148.58 , \text{K}\)
Convert to Celsius:
\(T_2 = 148.58 – 273.15\)
\(T_2 = -124.57^\circ \text{C}\)
Answer: The gas will have half the volume at approximately -125°C.
Question 10: Predict the final volume of CO₂ heated from 25°C to 190°C.
Given:
- Initial volume: \(V_1 = 0.10 , \text{L}\)
- Initial temperature: \(T_1 = 25^\circ \text{C} = 298.15 , \text{K}\)
- Final temperature: \(T_2 = 190^\circ \text{C} = 463.15 , \text{K}\)
- Final volume: \(V_2 = ?\)
Solution:
Using Charles’s Law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(V_2\):
\(V_2 = V_1 \cdot \frac{T_2}{T_1}\)Substitute the given values:
\(V_2 = 0.10 \cdot \frac{463.15}{298.15}\)
\(V_2 = 0.10 \cdot 1.553\)
\(V_2 = 0.155 , \text{L}\)
Answer: The final volume is approximately 0.16 L.
Question 11: What was the initial volume of gas at 50°C?
Given:
- Final volume: \(V_2 = 2.00 , \text{L}\)
- Final temperature: \(T_2 = 25.0^\circ \text{C} = 298.15 , \text{K}\)
- Initial temperature: \(T_1 = 50^\circ \text{C} = 323.15 , \text{K}\)
- Initial volume: \(V_1 = ?\)
Solution:
Using Charles’s Law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)Rearranging to solve for \(V_1\):
\(V_1 = V_2 \cdot \frac{T_1}{T_2}\)Substitute the given values:
\(V_1 = 2.00 \cdot \frac{323.15}{298.15}\)
\(V_1 = 2.00 \cdot 1.084\)
\(V_1 = 2.17 , \text{L}\)
Answer: The initial volume was 2.17 L.
Question 12: To what temperature must hydrogen gas be heated?
Given:
- Initial volume: \(V_1 = 45.0 , \text{L}\)
- Initial temperature: \(T_1 = 290 , \text{K}\)
- Final volume:
text{L}\)
- Final temperature: \(T_2 = ?\)
Solution:
Using Charles’s Law:
\(T_2 = T_1 \cdot \frac{V_2}{V_1}\)Substitute the given values:
\(T_2 = 290 \cdot \frac{65.0}{45.0}\)
\(T_2 = 290 \cdot 1.444\)
\(T_2 = 419 , \text{K}\)
Answer: The gas must be heated to 419 K.
Question 13: What was the initial temperature to achieve 146°C?
Given:
- Initial volume: \(V_1 = 45.0 , \text{L}\)
- Final volume: \(V_2 = 65.0 , \text{L}\)
- Final temperature: \(T_2 = 146^\circ \text{C} = 419 , \text{K}\)
- Initial temperature: \(T_1 = ?\)
Solution:
Using Charles’s Law:
\(T_1 = T_2 \cdot \frac{V_1}{V_2}\)Substitute the given values:
\(T_1 = 419 \cdot \frac{45.0}{65.0}\)
\(T_1 = 419 \cdot 0.692\)
\(T_1 = 290 , \text{K}\)
Convert to Celsius:
\(T_1 = 290 – 273.15\)
\(T_1 = 16.85^\circ \text{C}\)
Answer: The initial temperature was approximately 17°C.