The following question has troubled me for a while.

For gases that are slightly soluble in water, there is a proportional relationship between the partial pressure, $P$, and the mole fraction, $x$, of the gas molecules dissolved in water (Henry's law):

$\ce{P = k_H*x}$

A container ($V$ = $570$ $mL$) is filled with water ($V_l$ = $500$ $mL$) and pressurized with $CO_2$ gas ($P_0$ = $50$ $atm$), before it is allowed to stand at $10^\circ$C until the vapor–liquid equilibrium is established.

Calculate the pressure of $CO_2$ $[atm]$ in the container and the amount of $CO_2$ $[mol]$ dissolved in the water. The Henry coefficient of $CO_2$ for water at $10^\circ$C is $k_H$ = $0.104 × 10^4$ $atm$, and we will consider that the reaction of $CO_2$ in water can be ignored.

I did the following calculations but I am unsure with the outcome.

$\ce{n_{total}\approx n(H_2O) = \frac{\rho V}{M(H_2O)} = 27.753 mol}$

$\ce{x = \frac{P_0}{k_H} = 0.048077}$

$\ce{n(CO_2)_{dissolved} = x*n_{total} \approx 1.3343 mol}$

$\ce{P_1 = \frac{nRT}{V} = \frac {1.3343*0.0831451*1.01325*283}{0.570} = 55.8 atm}$

Shouldn't the final pressure be smaller than the initial pressure ($50$ $atm$)?

  • 2
    $\begingroup$ yes, the problem is wanting you to split the $\ce{CO2}$ initially introduced into the container between the gas and liquid phase. $\endgroup$
    – MaxW
    May 4 at 20:55


  • Reaction of carbon dioxide with water is neglected.
  • Vapour pressure of water is negligible.
  • Volume of solution does not change on dissolution of carbon dioxide.
  • Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2{(aq)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$

Initially no $\ce{CO2}$ is dissolved in water and so initial moles of $\ce{H2O{(l)}}$ and $\ce{CO2{(g)}}$ can be calculated as follows:

$$ n_{{\ce{H2O{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\ n_{{\ce{CO2{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\ $$

Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.

Final pressure in the container can be written as:

$$ P_f= \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\ $$

And by Henry's law we can say,

$$ P_f= K_H \times X_{\ce{CO2{(aq)}}} $$

Thus from both these equations we can conclude:

$$ K_H \times \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\ $$

We also known that,

$$ n_{{\ce{CO2{(g)}}}_f} = n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f} \\ $$


$$ K_H \times \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f})RT}{V}\\ $$

Solving this we get,

$$ n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\ $$

This correspondingly means:

$$ n_{{\ce{CO2{(g)}}}_f}= \pu{0.015 mol}\\ P_f= \pu{5 atm}\\ $$

The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $P_f=P_i$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.


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