Final answer:
Using the combined gas law to solve for the temperature when the volume of a gas changes to 78.31 L at a pressure of 612.0 mm Hg, results in an approximate temperature of 27°C, corresponding to option B).
Step-by-step explanation:
To determine at what temperature the gas has a volume of 78.31 L at a pressure of 612.0 mm Hg, the combined gas law (P1V1/T1 = P2V2/T2) can be utilized. Here, P1 and V1 represent the initial pressure and volume, T1 is the initial temperature, and P2, V2, and T2 are the final pressure, volume, and temperature respectively. At STP (Standard Temperature and Pressure), the conditions are 0°C or 273 K and 1 atm. The pressure at STP (1 atm) needs to be converted to mm Hg (1 atm = 760 mm Hg).
Since we're given the volume of gas at STP (62.65 L) where P1 = 760 mm Hg and T1 = 273 K, and the final volume V2 is 78.31 L with a pressure P2 of 612.0 mm Hg, we need to solve for T2 in Kelvin using the combined gas law. Once we solve for T2 in Kelvin, we convert it to °C by subtracting 273.
Calculating the temperature:
(760 mm Hg) × (62.65 L) / (273 K) = (612.0 mm Hg) × (78.31 L) / T2
Solving for T2 gives us:
T2 = (612.0 mm Hg × 78.31 L × 273 K) / (760 mm Hg × 62.65 L)
T2 = 300.95 K
Converting T2 to Celsius: T2 - 273 = 27.95°C
This means the approximate temperature would be 27°C, which corresponds to option B) 27°C.