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 92.5 ml bulb that contains 0.195 g of br2 vapor at a pressure of 0.516 atm?

Answer 1

The temperature of a 92.5 mL bulb containing 0.195 g bromine vapor at a pressure of 0.516 atm is approximately 47.507 K, calculated using the Ideal Gas Law and appropriate conversions for volume and molar mass.

Step-by-step explanation:

To find the temperature in Kelvin at which bromine (Br2) vapor exists in a 92.5 mL bulb at a pressure of 0.516 atm with a mass of 0.195 g, we can use the Ideal Gas Law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm / (mol K)), and T is the temperature in Kelvin.

First, we must convert the volume from mL to L by dividing by 1000, so 92.5 mL becomes 0.0925 L. Next, we calculate the number of moles (n) of Br2 by using its molar mass (159.808 g/mol). Moles of Br2 = 0.195 g / 159.808 g/mol = 0.00122 mol.

Inserting these values into the Ideal Gas Law, we can solve for T as follows:

(0.516 atm) (0.0925 L) = (0.00122 mol) (0.0821 L atm / (mol K)) T

T = (0.516 atm * 0.0925 L) / (0.00122 mol * 0.0821 L atm / (mol K))

T = 47.507 K

Therefore, the temperature of the bulb that contains 0.195 g of Br2 vapor at a pressure of 0.516 atm is approximately 47.507 K.