Question

A flask of fixed volume contains 1.0 mole of gaseous carbon dioxide and 88 g of solid carbon dioxide. The original pressure and temperature in the flask is 1.0 atm and 300. K. All of the solid carbon dioxide sublimes. The final pressure in the flask is 2.5 atm. What is the final temperature? Assume the solid carbon dioxide takes up negligible volume.A. 150 KB. 200 KC. 250 KD. 300 KE. 400 K

Answer 1

The final temperature of the flask: T₂= 250 K

Step-by-step explanation:

Given: Volume of gas: V = constant, mass of CO₂: w = 88 g

Initial pressure of gas: P₁ = 1 atm, Initial temperature: T₁ = 300 K, Initial number of moles of gas: n₁ = 1 mole

After sublimation- Final pressure of gas: P₂= 2.5 atm, Final temperature: T₂=?K

Molar mass of CO₂: m = 44 g/mol, given mass of CO₂: w = 88 g

Therefore, the final number of moles of CO₂ gas after sublimation of 88 g solid CO₂: n₂ = initial number of moles + number of moles sublimed

∴ n₂ = n₁ + (w ÷ m) = 1 mole + (88 g ÷ 44 g/mol) = 1 mole + 2 mole

→ n₂ = 3 moles

To find the final temperature of CO₂ gas, we use the Gay-Lussac's law:

`(P_(1))/(n_(1)* T_(1)) = (P_(2))/(n_(2)* T_(2))`

→
`T_(2) = (P_(2)* n_(1)* T_(1))/(P_(1)* n_(2))`

→
`T_(2) = (2.5 atm* 1 mol* 300 K)/( 1 atm* 3 mol)`

→
`T_(2) = 250 K`

Therefore, the final temperature of the flask: T₂= 250 K