I am currently studying gaseous state and I encountered the following problem.

A container with a volume 3 liter holds $\ce{N2(g)}$ and $\ce{H2O(l)}$ at $\pu{29 ^\circ C}$. The pressure is found to be $\pu{1 atm}$. The water is then split into hydrogen and oxygen by electrolysis according to the reaction $$\ce{H2O (l) -> H2 (g) + 1/2 O2 (g)}$$ After the reaction is complete, the pressure is $\pu{1.86 atm}$. What mass of water was present in the container? The aqueous tension of water at $\pu{29 ^\circ C}$ is $\pu{0.04 atm}$.

This is one of the steps of the solution given along with the question (edited to show the units):

Amount of substance of nitrogen: $$n(\ce{N2}) = \frac{\pu{0.96 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.116 mol}$$ Amount of substance of water: $$n(\ce{H2O}) = \frac{\pu{0.04 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.00484 mol}$$

The thing I don't understand here is that how can the volume of nitrogen and water be taken as $\pu{3 l}$? Both of them would occupy some volume not the whole volume will not be occupied by one of them.

  • $\begingroup$ Hello and welcome to our site! We have a chat room specifically for JEE Chemistry. You're welcome to join it, but you'll have to cross 20 reputation first to speak there. $\endgroup$ Jul 10 '19 at 4:34
  • $\begingroup$ You asked: "how can the volume of nitrogen and water be taken as 3 L?" The answer is that the water is primarily in the liquid phase (with some water vapor, at a pressure equal to the vapor pressure of water at 29C), while the nitrogen is in the gas phase. Thus you have liquid water at the bottom of the container, and gaseous nitrogen and gaseous water filling the remainder of the 3 L container, giving a total volume for all species of 3 L. $\endgroup$
    – theorist
    Jul 10 '19 at 4:56
  • $\begingroup$ @theorist That is where I am confused. After the water is filled, the volume left will not be 3 L. Instead, it will be less than 3 as some volume will be occupied by the water. Thus your statement "remainder of the 3 L container", doesn't really makes any semce. $\endgroup$ Jul 10 '19 at 4:58
  • $\begingroup$ Not sure about the solution, but after the reaction is complete, you've got a 3L container that contains a mixture of N2(g), H2(g), and O2(g). The volume occupied by each is 3L, since each gas has full access to the entire container. Consider how you would calculate n for each gas with the ideal gas law: n=PV/(RT). For each gas, you would use P = the partial pressure of the gas, and V = the volume of the container (3L). [Or, equivalently, you could use P = total pressure in the container (1.86 atm), and V = partial molar volume of the gas; but I'm guessing the solution used the former.] $\endgroup$
    – theorist
    Jul 10 '19 at 5:07
  • 1
    $\begingroup$ @theorist Please write an answer and avoid answering the question in comments itself! Have a great day. $\endgroup$ Jul 10 '19 at 7:14

The idea behind the question is that you know the partial pressure of N$_2$ is 0.96 atm (= 1 atm - 0.04 atm). And you know that the partial pressure of all the H$_2$ that can be produced from the water is thus (1.86 atm - 0.96 atm)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the number of moles of H$_2$, which is equal to the number of moles of water = 0.6 atm *3 L/(302 K * 0.0821 atm L / mol K).

Note the number of moles of water given in your answer isn't the total no. of moles of water, it's the number of moles of water vapor prior to the electrolysis.


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