Question

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature (in K) in a 71.4 mL bulb that contains 0.276 g of I2 vapor at a pressure of 0.478 atm?

Answer 1

382.63 K

Step-by-step explanation:

We are given;

• Volume of Iodine as 71.4 mL
• Mass of Iodine as 0.276 g
• Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

• We are going to use the ideal gas equation;

• According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.

• Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

= 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

= 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K