A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 °C. What is the final volume of the gas, in milliliters, when the pressure is changed to 0.80 atm and the temperature to 25 °C, assuming the amount of gas remains constant?

• A. 735 mL
• B. 900 mL
• C. 500 mL
• D. 1,200 mL

Answer: (D)

Verified step by step guidance

1. Identify the initial and final conditions of the gas: Initial volume (V1) = 735 mL, initial pressure (P1) = 1.20 atm, initial temperature (T1) = 112 °C, final pressure (P2) = 0.80 atm, and final temperature (T2) = 25 °C.
2. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each: T1 = 112 + 273.15 K and T2 = 25 + 273.15 K.
3. Use the combined gas law, which is \( \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \), to relate the initial and final states of the gas.
4. Rearrange the combined gas law to solve for the final volume (V2): \( V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{P_2 \cdot T_1} \).
5. Substitute the known values into the equation and solve for V2: \( V_2 = \frac{1.20 \text{ atm} \cdot 735 \text{ mL} \cdot (25 + 273.15) \text{ K}}{0.80 \text{ atm} \cdot (112 + 273.15) \text{ K}} \).