Answer:
The final temperature of the flask: T₂= 250 K
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:
[tex]\frac{P_{1}}{n_{1}\times T_{1}} = \frac{P_{2}}{n_{2}\times T_{2}}[/tex]
→ [tex]T_{2} = \frac{P_{2}\times n_{1}\times T_{1}}{P_{1}\times n_{2}}[/tex]
→ [tex]T_{2} = \frac{2.5 atm\times 1 mol\times 300 K}{ 1 atm\times 3 mol}[/tex]
→ [tex]T_{2} = 250 K[/tex]
Therefore, the final temperature of the flask: T₂= 250 K