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

In summary, the conversation discusses the use of values in a formula for calculating the system work in a process involving the dissociation of gases. The confusion arises from the use of 22.4x10^-3 instead of 24x10^-3 in the molar volume, and the division of 298K by 273K. The explanation is given that these values are used to calculate the molar volume at a different temperature, and this can be derived from the ideal gas law. The conversation also touches on the use of the real gas law in solving for work in water electrolysis.
  • #1
Nico123
8
0
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
 
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  • #2
Nico123 said:
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
$$
\begin{align}
V_\mathrm{gas} &= n V_{\mathrm{m},298\ \mathrm{K}} \\
&= (1.5\ \mathrm{moles}) (22.4 \times 10^{-3}\ \mathrm{m}^3 \mathrm{mol}^{-1} ) \left(\frac{298\ \mathrm{K}}{273\ \mathrm{K}} \right)
\end{align}
$$
 
  • #3
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
 
Last edited:
  • #4
Nico123 said:
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 :confused:
 
  • #5
It's a bit of extra work than I need to research and Do
 
  • #6
Also how to you get from PV=nRt to V1/V2=T1/T2??
 
  • #7
Nico123 said:
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.
 
  • #8
Ok, Thanks so much
 
  • #9
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?
 
  • #10
HelloCthulhu said:
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?
Don't tag on to an existing thread. Post your question (with more details) in a new thread.
 

1. What is electrolysis of water?

Electrolysis of water is a chemical process in which an electric current is passed through a water solution to separate it into its two components, hydrogen and oxygen.

2. How does electrolysis of water work?

During electrolysis of water, the water molecules are broken down into their constituent elements, hydrogen and oxygen, through the process of oxidation and reduction. The electric current causes an oxidation reaction at the anode, where oxygen gas is produced, and a reduction reaction at the cathode, where hydrogen gas is produced.

3. What is the purpose of electrolysis of water?

The main purpose of electrolysis of water is to produce hydrogen gas, which has a wide range of industrial and commercial applications. It can also be used to produce oxygen gas, which is used in medical and scientific fields.

4. What are the factors that affect the rate of electrolysis of water?

The rate of electrolysis of water is affected by several factors, including the concentration of dissolved ions in the water, the type of electrodes used, the current intensity, and the temperature. Higher concentrations of ions and higher current intensities can increase the rate of electrolysis, while lower temperatures can slow it down.

5. What are the potential hazards of electrolysis of water?

One of the main hazards of electrolysis of water is the production of hydrogen gas, which is highly flammable and can pose a fire or explosion risk if not handled properly. Additionally, the use of electricity and corrosive chemicals in the process can also present safety hazards if proper precautions are not taken.

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