5 G Of Water Is Electrolyzed At A Constant Pressure At $25^{\circ} C$. Calculate The Expansion Work Done By The System, Assuming It Is An Open System.

Introduction

Electrolysis is a process in which an electric current is used to drive a chemical reaction. In this process, water is split into hydrogen and oxygen gases. The electrolysis of water is an important process in various industries, including the production of hydrogen fuel cells and the purification of water. In this article, we will calculate the expansion work done by the system when 5 g of water is electrolyzed at a constant pressure at $25^{\circ} C$.

Understanding Expansion Work

Expansion work is the work done by a system as it expands against an external pressure. In the case of electrolysis, the system is the water, and the external pressure is the atmospheric pressure. As the water is split into hydrogen and oxygen gases, the system expands, and work is done against the external pressure.

Calculating Expansion Work

To calculate the expansion work done by the system, we need to use the following formula:

$$W = -P \Delta V$$

where $W$ is the work done, $P$ is the external pressure, and $\Delta V$ is the change in volume of the system.

Step 1: Calculate the Number of Moles of Water

First, we need to calculate the number of moles of water. The molecular weight of water is 18 g/mol. Therefore, the number of moles of water is:

$$n = \frac{m}{M} = \frac{5 \text{ g}}{18 \text{ g/mol}} = 0.278 \text{ mol}$$

Step 2: Calculate the Change in Volume

Next, we need to calculate the change in volume of the system. The volume of the system is the sum of the volumes of the hydrogen and oxygen gases. The volume of the hydrogen gas is:

$$V_H = \frac{n_H RT}{P}$$

where $n_H$ is the number of moles of hydrogen, $R$ is the gas constant, $T$ is the temperature in Kelvin, and $P$ is the external pressure.

The number of moles of hydrogen is half the number of moles of water, since each molecule of water produces one molecule of hydrogen:

$$n_H = \frac{n}{2} = \frac{0.278 \text{ mol}}{2} = 0.139 \text{ mol}$$

The volume of the hydrogen gas is:

$$V_H = \frac{0.139 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}}{1 \text{ atm}} = 3.33 \text{ L}$$

The volume of the oxygen gas is:

$$V_O = \frac{n_O RT}{P}$$

where $n_O$ is the number of moles of oxygen.

The number of moles of oxygen is half the number of moles of water, since each molecule of water produces one molecule of oxygen:

$$n_O = \frac{n}{2} = \frac{0.278 \text{ mol}}{2} = 0.139 \text{ mol}$$

The volume of the oxygen gas is:

$$V_O = \frac{0.139 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}}{1 \text{ atm}} = 3.33 \text{ L}$$

The total change in volume is the sum of the volumes of the hydrogen and oxygen gases:

$$\Delta V = V_H + V_O = 3.33 \text{ L} + 3.33 \text{ L} = 6.66 \text{ L}$$

Step 3: Calculate the Expansion Work

Finally, we can calculate the expansion work done by the system:

$$W = -P \Delta V = -1 \text{ atm} \times 6.66 \text{ L} = -6.66 \text{ L atm}$$

To convert this to joules, we can use the following conversion factor:

$$1 \text{ L atm} = 101.325 \text{ J}$$

Therefore, the expansion work done by the system is:

$$W = -6.66 \text{ L atm} \times 101.325 \text{ J/L atm} = -675.5 \text{ J}$$

Conclusion

In this article, we calculated the expansion work done by the system when 5 g of water is electrolyzed at a constant pressure at $25^{\circ} C$. The expansion work is a measure of the work done by the system as it expands against an external pressure. We used the formula $W = -P \Delta V$ to calculate the expansion work, where $P$ is the external pressure and $\Delta V$ is the change in volume of the system. The result shows that the expansion work done by the system is -675.5 J.

References

• [1] Atkins, P. W. (2018). Physical Chemistry. Oxford University Press.
• [2] Chang, R. (2015). Physical Chemistry for the Biosciences. University Science Books.
• [3] Levine, I. N. (2014). Physical Chemistry. McGraw-Hill Education.

Table of Contents

1. Introduction
2. Understanding Expansion Work
3. Calculating Expansion Work
4. Step 1: Calculate the Number of Moles of Water
5. Step 2: Calculate the Change in Volume
6. Step 3: Calculate the Expansion Work
7. Conclusion
8. References
9. Table of Contents
   5 g of Water is Electrolyzed: Q&A =====================================

Q: What is electrolysis?

A: Electrolysis is a process in which an electric current is used to drive a chemical reaction. In this process, water is split into hydrogen and oxygen gases.

Q: What is the purpose of electrolysis?

A: The purpose of electrolysis is to produce hydrogen gas, which can be used as a fuel source. Hydrogen gas is a clean-burning fuel that produces only water vapor and heat as byproducts.

Q: What are the products of electrolysis?

A: The products of electrolysis are hydrogen gas and oxygen gas.

Q: What is the reaction equation for electrolysis?

A: The reaction equation for electrolysis is:

$$2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$$

Q: What is the role of the anode and cathode in electrolysis?

A: The anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction occurs. In electrolysis, the anode is where oxygen gas is produced, and the cathode is where hydrogen gas is produced.

Q: What is the significance of the external pressure in electrolysis?

A: The external pressure is the pressure exerted by the surroundings on the system. In electrolysis, the external pressure is important because it affects the volume of the system and, therefore, the expansion work done by the system.

Q: How is the expansion work calculated in electrolysis?

A: The expansion work is calculated using the formula:

$$W = -P \Delta V$$

where $W$ is the work done, $P$ is the external pressure, and $\Delta V$ is the change in volume of the system.

Q: What is the relationship between the number of moles of water and the number of moles of hydrogen and oxygen?

A: The number of moles of hydrogen and oxygen is half the number of moles of water, since each molecule of water produces one molecule of hydrogen and one molecule of oxygen.

Q: What is the significance of the temperature in electrolysis?

A: The temperature is important in electrolysis because it affects the rate of the reaction and the volume of the system.

Q: Can electrolysis be used to purify water?

A: Yes, electrolysis can be used to purify water. By removing impurities from the water, electrolysis can produce clean water that is free from contaminants.

Q: What are the applications of electrolysis?

A: The applications of electrolysis include the production of hydrogen gas, the purification of water, and the production of oxygen gas.

Q: Is electrolysis a safe process?

A: Yes, electrolysis is a safe process when performed properly. However, it is essential to follow proper safety protocols to avoid accidents and injuries.

Q: Can electrolysis be used in industrial processes?

A: Yes, electrolysis can be used in industrial processes, such as the production of chemicals and the purification of water.

Q: What are the advantages of electrolysis?

A: The advantages of electrolysis include the production of clean-burning fuels, the purification of water, and the production of oxygen gas.

Q: What are the disadvantages of electrolysis?

A: The disadvantages of electrolysis include the high energy requirements, the need for specialized equipment, and the potential for accidents and injuries.

Table of Contents

1. Q: What is electrolysis?
2. Q: What is the purpose of electrolysis?
3. Q: What are the products of electrolysis?
4. Q: What is the reaction equation for electrolysis?
5. Q: What is the role of the anode and cathode in electrolysis?
6. Q: What is the significance of the external pressure in electrolysis?
7. Q: How is the expansion work calculated in electrolysis?
8. Q: What is the relationship between the number of moles of water and the number of moles of hydrogen and oxygen?
9. Q: What is the significance of the temperature in electrolysis?
10. Q: Can electrolysis be used to purify water?
11. Q: What are the applications of electrolysis?
12. Q: Is electrolysis a safe process?
13. Q: Can electrolysis be used in industrial processes?
14. Q: What are the advantages of electrolysis?
15. Q: What are the disadvantages of electrolysis?