Question

Bromine, Br2, is a liquid at room temperature but boils (converts from a liquid into a gas) when warmed. What is the temperature (in K) in a 83.1 mL bulb that contains 0.181 g of Br2 vapor at a pressure of 0.481 atm

Answer 1

The temperature within the 83.1 mL bulb containing 0.181 g of bromine vapor at a pressure of 0.481 atm is approximately 359 K, as calculated using the Ideal Gas Law formula.

Step-by-step explanation:

The question is asking to find the temperature in Kelvin inside a bulb with a volume of 83.1 mL that contains 0.181 g of bromine vapor at a pressure of 0.481 atm. To find this, we can use the Ideal Gas Law, which is formulated as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature in Kelvin.

The first step is to calculate the number of moles of bromine vapor using its molar mass (Br2 has a molar mass of approximately 159.808 g/mol).

So the moles of Br2 are calculated by 0.181 g / 159.808 g/mol = 0.00113 moles.

Next, we convert the volume from mL to L by dividing by 1000, giving us 0.0831 L.

The universal gas constant R is 0.0821 L·atm/K·mol.

We can rearrange the Ideal Gas Law to solve for T (temperature in Kelvin): T = PV / (nR).

Plugging our values into the equation, we get T = (0.481 atm × 0.0831 L) / (0.00113 moles × 0.0821 L·atm/K·mol) which calculates to a temperature of approximately 359 K.