Final answer:
Bromine (Br₂) is a liquid at room temperature, but it boils and converts into a gas when warmed. To calculate the temperature in Kelvin at which 0.217 g of Br₂ vapor at a pressure of 0.511 atm occupies a volume of 82.0 mL, we can use the ideal gas law equation. By determining the number of moles of Br₂ and plugging in the values for pressure, volume, number of moles, and the ideal gas constant, we can calculate the temperature to be approximately 925 K.
Step-by-step explanation:
At room temperature, bromine (Br₂) is a liquid. However, when warmed, it undergoes a process called vaporization, where it converts from a liquid to a gas. The temperature at which this conversion occurs is known as the boiling point of bromine.
In order to determine the temperature (in Kelvin) at which 0.217 g of Br₂ vapor at a pressure of 0.511 atm occupies a volume of 82.0 mL, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for temperature (T), we get T = PV / (nR), where P is the pressure, V is the volume, n is the number of moles, and R is the ideal gas constant. Given the pressure and volume, we need to first calculate the number of moles of Br₂ using its molar mass. Then, we can plug in the values to calculate the temperature in Kelvin.
Using the molar mass of bromine (Br₂) as 159.808 g/mol, we can find the number of moles:
moles = mass / molar mass = 0.217 g / 159.808 g/mol = 0.00136 mol
Plugging in the values for pressure (0.511 atm), volume (82.0 mL converted to L, i.e., 0.0820 L), the number of moles (0.00136 mol), and the ideal gas constant (0.0821 L·atm/mol·K), we can calculate the temperature:
T = (0.511 atm) * (0.0820 L) / (0.00136 mol * 0.0821 L·atm/mol·K) ≈ 925 K