Question

A 2 L container holds oxygen gas at 300 K and 2 atm pressure. If the temperature is increased to 600 K and the volume is doubled, what is the final pressure?

• A. 1 atm
• B. 2 atm
• C. 0.5 atm
• D. 4 atm

Hint:

Use the combined gas law \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) when pressure, volume, and temperature all change.

Answer 1

The Correct Option is B

Step 1: Write down the initial conditions: \[ P_1 = 2 \text{ atm}, \quad V_1 = 2 \text{ L}, \quad T_1 = 300 \text{ K} \]
Step 2: Write down the final conditions: \[ V_2 = 2 \times V_1 = 4 \text{ L}, \quad T_2 = 600 \text{ K}, \quad P_2 = ? \]
Step 3: Use the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Step 4: Substitute the known values: \[ \frac{2 \times 2}{300} = \frac{P_2 \times 4}{600} \] \[ \frac{4}{300} = \frac{4 P_2}{600} \implies \frac{4}{300} = \frac{4 P_2}{600} \]
Step 5: Simplify and solve for \( P_2 \): \[ \frac{4}{300} = \frac{4 P_2}{600} \implies \frac{4}{300} \times \frac{600}{4} = P_2 \implies \frac{600}{300} = P_2 \implies 2 = P_2 \]
Step 6: Check for calculation correctness.
Note: The previous calculation shows \(P_2 = 2\) atm, but let's carefully re-check the algebra. \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \] \[ P_2 = \frac{2 \times 2 \times 600}{300 \times 4} = \frac{2400}{1200} = 2 \text{ atm} \] Final answer: \(P_2 = 1\) atm or \(2\) atm? Actually, the volume is doubled and temperature doubled, so \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{2 \times 2 \times 600}{300 \times 4} = \frac{2400}{1200} = 2 \text{ atm} \] So \(P_2 = 2\) atm. Hence, the correct answer is (B) 2 atm.