Finding partial pressure of a gas in a mixture

Question

A mixture of argon and carbon dioxide was initially at a total pressure of $\pu{1.72 atm}$ in a volume of $\pu{1.72 L}$ at $\pu{17 ^\circ C}$. The carbon dioxide was then completely removed from the mixture by passing it through a solution of sodium hydroxide ($\ce{NaOH}$). The argon was collected in a $\pu{1.1 L}$ vessel and its pressure found to be $\pu{1.55 atm}$ at a temperature of $\pu{28 ^\circ C}$. What was the partial pressure of carbon dioxide in the original gas mixture? Assume that argon and carbon dioxide behave as ideal gases.

So, initially I found the moles of argon in in a $\pu{1.1 L}$ vessel which is $\pu{0.06896 moles}$

Then, i subtracted $\pu{1.72 L}$ from $\pu{1.1 L}$ in order to get a volume of carbon dioxide and used it in $PV = nRT$ and i got $\pu{0.04474 moles}$ of carbon dioxide .

Finally, I used the formula for partial pressure

$\frac{\text{moles of carbon dioxide} \times\text{total pressure}}{\text{moles of argon} + \text{moles of carbon dioxide}}$ and got $\pu{0.6771 atm}$.

However, my answer is not correct according to the textbook. Where did i make a mistake?

Answer 1

Your intial step is correct, finding the number of moles of argon in the 1.1 vessel. The next step however is incorrect. Because the pressure and temperature are both in different conditions for argon in the end and argon and carbon dioxide together in the beginning, you cnnot simply subtract the volume 1.72 - 1.1 L and then find the number of moles of carbon dioxide there. Remember, subtracting the volume and finding the moles of co2 would only work if the conditions are the same, but in this case, the pressures and temepratures are different. Thus, you can just simply go about finding the total n moles of Argon and carbon dioxide through PV=nrt.

So, 1.72 (1.72L) / 0.08206 x (273 + 17) = 0.1243 total moles. Subtract 0.06869 moles from earlier from 0.1243. You will get 0.055356 moles of CO2. Now, you can use your partial pressure method.

Hope this helps.