For water at 100°C and 1 bar, Δ_vap H - Δ_vap U = ..... × 10^2 J mol^-1. (Round off to the nearest i

Q: Why do we ignore the volume of liquid water?

The volume of a gas is typically about 1000 times larger than the volume of the same mass of liquid. Therefore, $V_{\text{gas}} - V_{\text{liquid}} \approx V_{\text{gas}}$, making the liquid's volume negligible in the work calculation.

Q: Why is $\Delta n_g$ equal to 1?

In the reaction $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$, there is 1 mole of gas on the product side and 0 moles of gas on the reactant side. So, $1 - 0 = 1$.

Q: What if the process was condensation instead of vaporization?

For condensation, $\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l)$, $\Delta n_g$ would be $0 - 1 = -1$. The work would be done *on* the system, and $\Delta H$ would be more negative than $\Delta U$.

Q: Why must the temperature be in Kelvin?

The ideal gas constant $R$ is given in units of $\text{J mol}^{-1}\text{ K}^{-1}$. To ensure units cancel correctly, the temperature must always be in Kelvin.

Q: Is $\Delta H$ always greater than $\Delta U$ for vaporization?

Yes, because vaporization always involves an expansion against the surrounding pressure ($\Delta n_g > 0$), meaning the system must absorb extra heat to do this expansion work on top of increasing its internal energy.

Question

For water at $100^\circ	ext{C}$ and $1	ext{ bar}$, $\Delta_{	ext{vap}}H - \Delta_{	ext{vap}}U = \dots 	imes 10^2	ext{ J mol}^{-1}$.
(Round off to the nearest integer)
[Use : $R = 8.31	ext{ J mol}^{-1}	ext{ K}^{-1}$]
[Assume volume of $	ext{H}_2	ext{O}(l)$ is much smaller than volume of $	ext{H}_2	ext{O}(g)$. Assume $	ext{H}_2	ext{O}(g)$ treated as an ideal gas]

Sigma Percentile · Accepted Answer

Have you ever watched a pot of water boil and wondered about the invisible forces at play? It is a mesmerizing process where liquid water transforms into a gas, expanding and pushing against the atmosphere.

This simple everyday phenomenon is a perfect playground for the laws of thermodynamics. Let's dive deep into the physics and math behind the vaporization of water.

### Analyzing the Setup

When water turns from a liquid to a gas, it undergoes a profound transformation. It absorbs a significant amount of heat from its surroundings.

Because this process typically happens in an open container, the pressure remains constant at $1 	ext{ bar}$. The total heat absorbed under these conditions is known as the **enthalpy of vaporization**, denoted by $\Delta_{	ext{vap}}H$.

But where does all this energy go? It serves two distinct purposes.

First, it increases the internal energy of the water molecules, breaking the intermolecular hydrogen bonds and allowing them to move freely. This is the **change in internal energy**, $\Delta_{	ext{vap}}U$.

Second, as the water turns into steam, it expands massively. This expansion requires the system to do work against the surrounding atmospheric pressure.

### The Master Equation

The First Law of Thermodynamics beautifully connects these quantities. It is essentially the law of conservation of energy applied to thermodynamic systems.

The mathematical statement is:

$$ \Delta H = \Delta U + p\Delta V $$

Here, $p\Delta V$ represents the expansion work done by the gas.

Since we are dealing with water vapor, which we can approximate as an ideal gas, we can use the ideal gas law to rewrite this work term. For a process at constant temperature and pressure, the change in volume is directly proportional to the change in the number of moles of gas.

Therefore, we can substitute $p\Delta V$ with $\Delta n_g RT$:

$$ \Delta H = \Delta U + \Delta n_g RT $$

Rearranging this to find the difference between enthalpy and internal energy, we get:

$$ \Delta H - \Delta U = \Delta n_g RT $$

This elegant equation tells us that the difference between the heat supplied and the internal energy change is exactly equal to the expansion work done by the newly formed gas!

### Calculating the Gaseous Moles

To use our master equation, we first need to determine $\Delta n_g$, the change in the number of gaseous moles. Let's write down the balanced chemical equation for the vaporization of water.

$$ 	ext{H}_2	ext{O}(l) ightleftharpoons 	ext{H}_2	ext{O}(g) $$

The term $\Delta n_g$ is calculated by subtracting the total moles of gaseous reactants from the total moles of gaseous products.

$$ \Delta n_g = n_{	ext{gas, products}} - n_{	ext{gas, reactants}} $$

Looking at our equation, we have $1 	ext{ mole}$ of water vapor on the product side and $0 	ext{ moles}$ of gas on the reactant side (since liquid water doesn't count).

$$ \Delta n_g = 1 - 0 = 1 $$

This means for every mole of water vaporized, exactly one mole of gas is created, driving the expansion work.

### Final Calculation

Now we have all the pieces of the puzzle. We can substitute our known values into the rearranged First Law equation.

We know $\Delta n_g = 1$, the ideal gas constant $R = 8.31 	ext{ J mol}^{-1}	ext{ K}^{-1}$, and the boiling temperature $T = 100^\circ	ext{C}$.

**Always remember to convert temperature to Kelvin in thermodynamics!** So, $T = 100 + 273 = 373 	ext{ K}$.

Let's plug these into our equation:

$$ \Delta H - \Delta U = (1) 	imes (8.31) 	imes (373) $$

Multiplying these numbers together gives us the raw energy difference.

$$ \Delta H - \Delta U = 3099.63 	ext{ J mol}^{-1} $$

The question asks us to format this answer as a number multiplied by $10^2$. To do this, we shift the decimal point two places to the left.

$$ \Delta H - \Delta U = 30.9963 	imes 10^2 	ext{ J mol}^{-1} $$

Finally, we need to round this value to the nearest integer. The decimal $.9963$ is very close to $1$, so we round up.

$$ \Delta H - \Delta U \approx \mathbf{31 	imes 10^2 	ext{ J mol}^{-1}} $$

### Conclusion

The final answer is **31**.

This problem perfectly illustrates how energy is partitioned during a phase change. The heat you supply to boil water doesn't just heat it up; a significant chunk of it is spent physically pushing the atmosphere out of the way to make room for the steam.

Understanding this balance is the key to mastering chemical thermodynamics!

Enter Numerical Value:

Visualized Solution

$Vaporization of Water$

$First Law of Thermodynamics$

$Δ n_{g} Calculation$

$Substitution$

$Final Calculation$

$Conclusion$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Calculating the Gaseous Moles

Final Calculation

Conclusion

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