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 K
B. 200 K
C. 250 K
D. 300 K
E. 400 K

Answer 1

Answer:

The correct option is: C. 250 K

Explanation:

Given: Before Sublimation-

Initial Temperature: T₁ = 300 K, Initial Pressure: P₁ = 1 atm, Initial number of moles of gas: n₁ = 1 mol, given mass of solid Carbon dioxide: w = 88 g    

After Sublimation-      

Final Pressure: P₂ = 2.5 atm, Final number of moles of gas: n₂ = ? mol

Final Temperature: T₂ = ? K,            

Also, Volume is constant, Molar mass of Carbon dioxide: m = 44 g/mol

As we know,

The number of moles:

$n = \frac {given\: mass\: (w)} {Molar\: mass\: (m)}$

So the number of moles of carbon dioxide sublimed:

$n = \frac {w}{m} = \frac {88\: g} {44\: g/mol} = 2 mol$

Therefore, the final number of moles of gas after sublimation:

$n_{2} = n_{1} + n = 1\: mol + 2\: mol = 3\: mol$

According to the Ideal gas equation:

$P.V = n.R.T$

$or, \frac {P_{1}.V_{1}}{n_{1}.T_{1}} = \frac {P_{2}.V_{2}}{n_{2}.T_{2}} \: \: \: \: \: \: ....equation\: (1)$

Since the volume is constant, so the equation (1) can be written as:

$\frac {P_{1}}{n_{1}.T_{1}} = \frac {P_{2}}{n_{2}.T_{2}}$

$\Rightarrow \frac {1\:atm}{1\:mol \times 300\:K} = \frac {2.5\:atm}{3\:mol \times T_{2}}$

$\therefore T_{2} = \frac {2.5\:atm \times 300\:K \times 1\:mol}{3\:mol \times 1\:atm}$

$\Rightarrow T_{2} = 250\:K$

Therefore, the final temperature: T₂ = 250 K