A sample of oxygen gas has a volume of 640 mL at a temperature of 23°C. At what temperature (in °C) will the volume of the oxygen be 820 mL, assuming pressure and amount of gas are constant?

• A. 60°C
• B. 35°C
• C. 70°C
• D. 50°C

Answer: (D)

Verified step by step guidance

1. First, identify the gas law that applies to this problem. Since pressure and the amount of gas are constant, we use Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin. The formula is \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
2. Convert the initial temperature from Celsius to Kelvin. The formula for conversion is \( T(K) = T(°C) + 273.15 \). So, \( T_1 = 23 + 273.15 \).
3. Set up the equation using Charles's Law: \( \frac{640 \text{ mL}}{T_1} = \frac{820 \text{ mL}}{T_2} \).
4. Solve for \( T_2 \) in Kelvin by rearranging the equation: \( T_2 = \frac{820 \text{ mL} \times T_1}{640 \text{ mL}} \).
5. Convert \( T_2 \) from Kelvin back to Celsius using the conversion formula \( T(°C) = T(K) - 273.15 \) to find the final temperature in Celsius.