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 a 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm?

a) 280 K
b) 325 K
c) 350 K
d) 392 K

Answer 1

The temperature in the 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm is 280 K. Option a

Step-by-step explanation:

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. To determine the temperature at which the given amount of I2 vapor exists, we can use the ideal gas law equation:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

We can rearrange the equation to solve for T:

T = PV / (nR)

Substituting the given values:

T = (0.462 atm) * (0.0733 L) / ((0.292 g) / (253.8 g/mol))

T = 280 K

Therefore, the temperature in the 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm is 280 K. Option a