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

85.7°C

Explanation:

Step 1:

Data obtained from the question. This include the following:

Volume (V) = 73.3mL = 73.3/1000 = 0.0733L

Mass of l2 = 0.292g

Pressure (P) = 0.462atm

Temperature (T) =..?

Step 2:

Determination of the number of mole present in 0.292g of I2. This is illustrated below:

Mass of l2 = 0.292g

Molar Mass of I2 = 127 x 2 = 254g/mol

Number of mole of I2 =..?

Number of mole = Mass /Molar Mass

Number of mole of I2 = 0.292/254

Number of mole of I2 = 1.15×10¯³ mole

Step 3:

Determination of the temperature.

The temperature in the bulb containing the iodine vapor can be obtained by using the ideal gas equation as follow:

Volume (V) = 0.0733L

Pressure (P) = 0.462atm

Number of mole (n) = 1.15×10¯³ mole

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =..?

PV = nRT

0.462 x 0.0733 = 1.15×10¯³ x 0.0821 x T

Divide both side by 1.15×10¯³ x 0.0821

T = (0.462 x 0.0733)/(1.15×10¯³ x 0.0821)

T = 358.7K

Converting 358.7K to celsius temperature, we have:

T(°C) = T(K) – 273

T(K) = 358.7K

T(°C) = 358.7 – 273

T(°C) = 85.7°C

Therefore, the temperature is 85.7°C