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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 83.3-mL bulb that contains 0.392 g of I2 vapor at a pressure of 0.562 atm

Respuesta :

Answer:

T = 3.75 K

Explanation:

As we know

PV = nRT

R =8.3144598 J. mol-1. K-1

P = 0.562 atm

V = 83.3 mL

moles in 0.392 g of I2 = 0.392/mass of I2 = 0.392 grams/253.8089 g/mol = 0.0015 moles

Substituting the given values, we get

0.562 atm * 83.3 *10^-3 L = 0.0015 moles * 8.3144598 J. mol-1. K-1 * T

T = 3.75 K

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