Final answer:
To find the temperature (in K) of bromine gas in a bulb at a given pressure and mass, we use the Ideal Gas Law. After converting the mass of Br2 to moles and the volume to liters, we find that the temperature of the gas is approximately 576.69 Kelvin.
Step-by-step explanation:
The question asks to determine the temperature in Kelvin of a 83.4 mL bulb containing 0.197 g of bromine gas at a pressure of 0.572 atm. To solve this, we use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. First, we convert the mass of bromine to moles using the molar mass of Br2 (159.808 g/mol).
To find the number of moles (n), we divide the mass of Br2 by its molar mass: n = 0.197 g / 159.808 g/mol = 0.001233 mol. Next, we convert the volume to liters by dividing by 1000: V = 83.4 mL / 1000 = 0.0834 L. Now we can rearrange the Ideal Gas Law to solve for T: T = (PV)/(nR). Plugging in the values we have T = (0.572 atm × 0.0834 L)/(0.001233 mol × 0.0821 L·atm/(mol·K)) = 576.69 K.
Therefore, the temperature of the bromine gas in the bulb is approximately 576.69 Kelvin.