A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas (gamma = 5/

Q: Why do we use $C_V$ for internal energy even though the process is happening at constant pressure?

The internal energy of an ideal gas is a state function, meaning it depends only on the current state (specifically, the temperature), not the path taken to get there. The formula $\Delta U = nC_V\Delta T$ is a universal truth for ideal gases across *any* process, whether it's constant volume, constant pressure, or adiabatic.

Q: How did we know the $C_V$ values for monatomic and diatomic gases without them being given?

These are standard values derived from the kinetic theory of gases. $C_V = \frac{f}{2}R$, where $f$ is the degrees of freedom. Monatomic gases have $3$ translational degrees of freedom ($f=3$), so $C_V = \frac{3}{2}R$. Diatomic gases have $3$ translational and $2$ rotational degrees of freedom ($f=5$), so $C_V = \frac{5}{2}R$.

Q: Could we have used the First Law of Thermodynamics ($\Delta Q = \Delta U + W$) to solve this?

Yes, absolutely! But it would take an extra step. You would first need to calculate the heat supplied $\Delta Q = nC_{P,\text{mix}}\Delta T$, which requires finding $C_{P,\text{mix}}$. Using $\Delta U = nC_{V,\text{mix}}\Delta T$ directly is a much faster and more elegant approach.

Q: Why is the total number of moles used in the work formula $W = n_{\text{total}} R \Delta T$?

In a gas mixture, the total pressure is the sum of the partial pressures of each gas. The total work done by the mixture is simply the sum of the work done by each individual gas expanding. Mathematically, $W = W_1 + W_2 = n_1 R \Delta T + n_2 R \Delta T = (n_1 + n_2) R \Delta T$.

Q: What if the gases were real instead of ideal?

If the gases were real, their internal energy would depend on both temperature and volume due to intermolecular forces. The simple formula $\Delta U = nC_V\Delta T$ would no longer be perfectly accurate, and we would need to use more complex equations of state like the Van der Waals equation.

Question

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas ($\gamma = 5/3$) and one mole of an ideal diatomic gas ($\gamma = 7/5$). Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of $66	ext{ Joule}$ when heated at constant pressure. The change in its internal energy is ________ Joule.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Gases

Imagine you are observing a closed container fitted with a movable piston. Inside this container, a fascinating thermodynamic dance is about to occur. We don't just have one gas; we have a homogeneous mixture of two distinct ideal gases.

Specifically, we have $n_1 = 2$ moles of a monatomic gas (like Helium or Neon) and $n_2 = 1$ mole of a diatomic gas (like Oxygen or Nitrogen). We are told that when this mixture is heated at a **constant pressure**, it expands and does a work of $W = 66	ext{ J}$. Our mission is to uncover the hidden change in the system's internal energy, $\Delta U$.

### The Master Equations

Before we dive into the numbers, let's lay out our mathematical tools. The work done by any ideal gas (or mixture) during an isobaric (constant pressure) process is given by:

$$W = n_{	ext{total}} R \Delta T$$

We know this value is exactly $66	ext{ J}$.

On the other hand, the change in internal energy for *any* process is strictly dependent on the temperature change and the molar heat capacity at constant volume, $C_V$. For our mixture, the equation is:

$$\Delta U = n_{	ext{total}} C_{V,	ext{mix}} \Delta T$$

### Finding the Mixture's Identity

To use our internal energy equation, we first need to figure out the equivalent heat capacity of our gas mixture, $C_{V,	ext{mix}}$. The formula for a mixture is simply the weighted average of the individual heat capacities:

$$C_{V,	ext{mix}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$$

From the kinetic theory of gases, we know that for a monatomic gas, $C_{V1} = \frac{3}{2}R$. For a diatomic gas, $C_{V2} = \frac{5}{2}R$. Let's substitute our known mole quantities into the formula:

$$C_{V,	ext{mix}} = \frac{2 \left(\frac{3}{2}Right) + 1 \left(\frac{5}{2}Right)}{2 + 1}$$

Simplifying the numerator, the $2$s cancel out in the first term, leaving us with $3R$. Adding this to $\frac{5}{2}R$ gives $\frac{11}{2}R$. Dividing by the total number of moles ($3$), we find our mixture's identity:

$$C_{V,	ext{mix}} = \frac{11}{6}R$$

### The Elegant Substitution

Now, let's bring this back to our internal energy equation. Substituting $C_{V,	ext{mix}}$ gives:

$$\Delta U = n_{	ext{total}} \left(\frac{11}{6}Right) \Delta T$$

Here is where the magic happens. Instead of trying to find the individual values of $n_{	ext{total}}$, $R$, or $\Delta T$, let's rearrange the terms slightly:

$$\Delta U = \frac{11}{6} (n_{	ext{total}} R \Delta T)$$

Do you recognize the term in the parentheses? It is exactly the expression for the work done at constant pressure! We already know that $n_{	ext{total}} R \Delta T = W = 66	ext{ J}$.

### The Final Reveal

By making this elegant substitution, we bypass a lot of unnecessary calculations. We simply plug in the value of the work done:

$$\Delta U = \frac{11}{6} 	imes 66$$

The math is beautifully simple. Six goes into $66$ exactly $11$ times. And $11$ multiplied by $11$ is $121$.

$$\Delta U = 121	ext{ J}$$

The change in the internal energy of our gas mixture is $121	ext{ J}$. By understanding the deep connections between work, internal energy, and heat capacity, we turned a potentially complex problem into a smooth, logical sequence.

Enter Numerical Value:

Visualized Solution

\text{System Setup}

\text{Work and Internal Energy}

C_{V,\text{mix}} \text{ Formula}

\text{Substituting Values}

\text{Calculating } C_{V,\text{mix}}

\text{Internal Energy Expression}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: First Law of Thermodynamics

The Setup

The Master Equations

Finding the Mixture's Identity

The Elegant Substitution

The Final Reveal

Similar Questions

For an ideal gas

Comprehension Passage

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

A system does $200 J$ of work and at the same time absorbs $150 J$ of heat. The magnitude of the change in internal energy is ......... $J$ . (Nearest integer)

A diatomic gas having $C_{p} = \frac{7}{2} R$ and $C_{V} = \frac{5}{2} R$ , is heated at constant pressure. The ratio $d U : d Q : d W$ is

At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ............

Five moles of an ideal gas at $1 bar$ and $298 K$ is expanded into vacuum to double the volume. The work done is

An ideal gas expands in volume from $1 im es 1 0^{- 3} m^{3}$ to $1 im es 1 0^{- 2} m^{3}$ at $300 K$ against a constant pressure of $1 im es 1 0^{5} Nm^{- 2}$ . The work done is