Internal Energy, Enthalpy, and Specific Heats for Ideal Gases

The Ideal Gas Superpower: Why Internal Energy Only Cares About Temperature

Let's start by revisiting a point from last time that might have felt a bit like a magic trick. We derived $du=c_{v}dT$ by looking at a constant volume process. But then I said you can use this formula for any process involving an ideal gas. How can that be?

The answer lies in a fundamental property of ideal gases: For an ideal gas, its internal energy is only a function of its temperature.

Think about it this way. What is internal energy at the molecular level? For an ideal gas, it's almost entirely the kinetic energy of its molecules—how fast they are zipping around and bouncing off each other. Temperature is just our macroscopic measure of this average molecular kinetic energy. So, it makes intuitive sense that if you raise the temperature, you raise the internal energy, and if you lower the temperature, you lower the internal energy, regardless of what the pressure or volume are doing.

Imagine your happiness level is your "internal energy." For an ideal gas, its happiness depends only on how much pizza (temperature) it has. It doesn't care if it's eating the pizza in a small room (constant volume) or a big park (constant pressure). If it goes from one slice to three slices, its happiness increases by the same amount in both scenarios.

This isn't just a convenient idea; it's a physical property that was confirmed experimentally.

To get a great visual explanation of this core concept and the experiment that proved it, let's watch a segment from the video we're using for this module.

So, to summarize the key takeaway from the video:

1. Experiments showed that when an ideal gas expands into a vacuum (doing no work and with no heat transfer), its temperature doesn't change.
2. According to the First Law ($\Delta U=Q-W$), if $Q=0$ and $W=0$, then $\Delta U$ must be 0.
3. Even though the gas's volume and pressure changed dramatically, its internal energy didn't change as long as the temperature was constant.
4. The conclusion: Internal energy $u$ is only a function of temperature $T$. We write this as $u=u(T)$.

And what about enthalpy, $h$?
We know $h=u+Pv$. For an ideal gas, we also know $Pv=RT$.
So, $h=u(T)+RT$. Since $u$ depends only on $T$, and $R$ is a constant, it must be that enthalpy $h$ is also only a function of temperature $T$. We write this as $h=h(T)$.

This is the "superpower." Because $u$ and $h$ are state functions that only depend on temperature for an ideal gas, the path taken between two temperatures doesn't matter. The change $\Delta u$ and $\Delta h$ will always be the same if the $\Delta T$ is the same.

This is why we are fully justified in using these equations for ANY process involving an ideal gas:
$\Delta u=u_2-u_1={\int }_{T_1}^{T_2}{c}_v(T)dT$
$\Delta h=h_2-h_1={\int }_{T_1}^{T_2}{c}_p(T)dT$

Recap Break

This was the most important conceptual part of the lesson. Let's lock it in.

• For an ideal gas, internal energy ($u$) and enthalpy ($h$) are functions of temperature only.
• This means that to find the change in internal energy or enthalpy, all you need to know is the change in temperature.
• The process path (constant pressure, constant volume, or something else) is irrelevant for calculating $\Delta u$ and $\Delta h$.
• This is why the formulas $\Delta u=c_v\Delta T$ and $\Delta h=c_p\Delta T$ are so universally useful for ideal gases.

A Real-World Wrinkle: Specific Heats Aren't Always Constant

So far, we've been using single values for $c_p$ and $c_v$ (like 1.005 kJ/kg·K for air). This is often called the cold-air-standard assumption, where we assume the specific heats are constant at their room temperature values. This is a good approximation for small temperature changes, but it can be inaccurate for large ones.

Why do specific heats change? Think about the molecules again. When we add energy (heat), it increases the kinetic energy of the gas molecules.

• At low temperatures, this energy mainly goes into making the molecules move faster (translational energy).
• At higher temperatures, the molecules start to vibrate and rotate more intensely as well. These are new "storage modes" for the energy.

Because it takes extra energy to get these vibrations and rotations going, more energy is required to raise the temperature by one degree. Therefore, the specific heats of most gases increase with temperature.

So, how do we accurately calculate $\Delta u$ and $\Delta h$ when $c_v$ and $c_p$ are changing over our temperature range? We have three common methods, ranging from most accurate to most approximate.

The next video segment explains these methods perfectly.

Let's summarize those three methods:

Method 1: Use Ideal Gas Property Tables (Most Accurate and Easiest)
For common gases like air, steam, CO₂, etc., engineers have already done the hard work. They have calculated the integrals $\int c_p(T)dT$ from a standard reference temperature (like 0 K) and tabulated the results.

• You will find tables (like Table A-17 for Air in many textbooks) that list internal energy $u$ and enthalpy $h$ for various temperatures.
• To find the change, you simply look up the values at your start and end temperatures and subtract:
  $\Delta h=h_2-h_1(\text{values from table})$
  $\Delta u=u_2-u_1(\text{values from table})$
• This method automatically accounts for the variation of specific heat with temperature and is the preferred method when tables are available.

Method 2: Use an Average Specific Heat (Good Approximation)
If you don't have tables or want a quicker calculation, you can approximate.

1. Find the specific heat at the initial temperature, $c_{p,1}$ at $T_1$.
2. Find the specific heat at the final temperature, $c_{p,2}$ at $T_2$.
3. Calculate the average specific heat: $c_{p,avg}=(c_{p,1}+c_{p,2})/2$.
4. Use this average value as if it were constant:
   $\Delta h\approx c_{p,avg}(T_2-T_1)$
   $\Delta u\approx c_{v,avg}(T_2-T_1)$
   This is much more accurate than just using the room temperature value, especially for large temperature changes.

Method 3: Assume Constant Specific Heats (Simplest Approximation)
This is the "cold-air-standard" method we started with. You assume $c_p$ and $c_v$ are constant at some known value (usually room temperature, ~300 K).
$\Delta h\approx c_p(T_2-T_1)$
$\Delta u\approx c_v(T_2-T_1)$
This is fine for back-of-the-envelope calculations or when the temperature change is small (less than a couple hundred degrees). But be aware of the potential for error!

Worked Numerical Problems

Let's see the difference these methods make in a real problem.
Consider heating 1 kg of air from $T_1=300\text{ K}$ (about 27°C) to $T_2=1000\text{ K}$ (about 727°C). We want to find the change in internal energy, $\Delta u$.

Given data:

• From an ideal gas table for air:
  • At $T_1=300\text{ K}$, $u_1=214.07\text{ kJ/kg}$, $c_v=0.718\text{ kJ/kg⋅K}$
  • At $T_2=1000\text{ K}$, $u_2=758.94\text{ kJ/kg}$, $c_v=0.855\text{ kJ/kg⋅K}$

Problem 1: Compare the three methods
Calculate $\Delta u$ using all three methods.

Problem 2: Proving the Path Doesn't Matter
A piston-cylinder contains 0.5 kg of helium. It undergoes a process from an initial state of 150 kPa and 20°C to a final state of 400 kPa and 100°C. For helium, $c_v$ is constant at 3.116 kJ/kg·K. Find the change in internal energy, $\Delta U$.

Conclusion

Great job today. We've taken a huge step forward in making our thermodynamic analysis both more powerful and more realistic.

Key Takeaways:

• The most important property of an ideal gas is that its internal energy ($u$) and enthalpy ($h$) are functions of temperature alone.
• This allows us to calculate $\Delta u$ and $\Delta h$ using $\Delta T$ for any process, not just constant volume or constant pressure ones.
• Specific heats ($c_p$ and $c_v$) are not truly constant; they generally increase with temperature.
• We have three methods to handle this:
  1. Property Tables: Easiest and most accurate.
  2. Average Specific Heat: A good approximation for larger temperature ranges.
  3. Constant Specific Heat (Cold Air Standard): Simplest, but only accurate for small temperature changes.

Preview of the Next Lesson

In our next two lessons, we're going to put everything from this module together. We'll start by systematically analyzing the four fundamental ideal gas processes:

1. Constant Volume (Isochoric)
2. Constant Pressure (Isobaric)
3. Constant Temperature (Isothermal)
4. No Heat Transfer (Adiabatic)

For each one, we now know how to calculate $\Delta U$ and $\Delta H$. The next step is to derive expressions for the work ($W$) and heat transfer ($Q$) for each process, allowing us to complete our first law energy balance ($Q-W=\Delta U$) for a full range of scenarios. See you then