How can the work for water electrolysis be solved using the real gas law?

[Nico123]

Hey Everyone,

I am a little confused about some of the values used in the following formula.

The process must provide the energy for the dissociation plus the energy to expand the produced gases. At temperature 298K and one atmosphere pressure, the system work is:

W = PΔV = (101.3 x 10^3 Pa)(1.5 moles)(22.4 x 10^-3 m3/mol)(298K/273K) = 3715 J

I understand why it is 101.3x10^3 Pa and also why it is 1.5moles, as it is H2 + 1/2 O2, but why is it 22.4x10^-3, when it is RTP. I though it should be 24x10^-3.
Also why do we do 298K/273K. I don't quite understand it.
Can someone please explain to me why these steps are done, and the reasoning behind it,
Thanks

 

[DrClaude]

but why is it 22.4x10^-3, when it is RTP. I though it should be 24x10^-3.
Also why do we do 298K/273K.

These two things go together. You seem to get that in the equation $\Delta V = V_f - V_i \approx V_\mathrm{gas} - 0$, with $V_\mathrm{gas}$ the amount of gas produced, which can be written as $n V_\mathrm{m}$, where $V_\mathrm{m}$ is the molar volume. As you mention, the molar volume of an ideal gas is known to be $\approx 22.4\ \mathrm{l}$ at STP, so you can calculate the molar volume at any temperature by using the ideal gas law $P V_\mathrm{m} = R T$:
$$
\frac{V_{\mathrm{m},T_2}}{V_{\mathrm{m},T_1}} = \frac{T_2}{T_1}
$$
You therefore have that
$$
V_{\mathrm{m},298\ \mathrm{K}} = V_{\mathrm{m},\mathrm{STP}} \frac{298\ \mathrm{K}}{273\ \mathrm{K}}
$$
hence
$$
\begin{align}
V_\mathrm{gas} &= n V_{\mathrm{m},298\ \mathrm{K}} \\
&= (1.5\ \mathrm{moles}) (22.4 \times 10^{-3}\ \mathrm{m}^3 \mathrm{mol}^{-1} ) \left(\frac{298\ \mathrm{K}}{273\ \mathrm{K}} \right)
\end{align}
$$

 

[Nico123]

Thank you so much, Just a quick doubt, is P=Pressure in the above equations and what is T and R? Also how to you get from PV=nRt to V1/V2=T1/T2

 

[DrClaude]

Thank you so much, Just a quick doubt, is P=Pressure in the above equations and what is T and R?

$T$ is temperature and $R$ is the gas constant.

I must say that I was taken aback by your question. It's hard for me to understand how you can be calculating things such as $W=P \Delta V$ and know about the molar volume of an ideal gas without having even learned the ideal gas law

 

[Nico123]

It's a bit of extra work than I need to research and Do

 

[Nico123]

Also how to you get from PV=nRt to V1/V2=T1/T2??

 

[DrClaude]

Also how to you get from PV=nRt to V1/V2=T1/T2??

You have P₁V₁ = n₁RT₁ and P₂V₂ = n₂RT₂. Pressure and number of moles are constant, P₁=P₂, n₁=n₂, so when you divide the equation 2 by equation 1, you get V₂/V₁ = T₂/T₁.

 

[Nico123]

Ok, Thanks so much

 

[HelloCthulhu]

Great explanation of how ideal gas law works in electrolysis. Can you explain how the work for water electrolysis could be solved using real gas law?

 

[DrClaude]

Great explanation of how ideal gas law works in electrolysis. Can you explain how the work for water electrolysis could be solved using real gas law?

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