A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at

Q: Why do we use $C_V$ and not $C_P$ here?

The internal energy of an ideal gas is always a function of temperature and is given by $n C_V T$, regardless of the thermodynamic process taking place. $C_P$ is used when calculating heat exchange at constant pressure, not internal energy.

Q: Does the volume of the vessel matter?

No, as long as the vessel is rigid (no work is done by the gas on the surroundings) and insulated, the volume doesn't affect the conservation of internal energy equation.

Q: What if the gases were different, like mixing Helium and Oxygen?

If the gases were different (e.g., monatomic and diatomic), their $C_V$ values would be different and wouldn't cancel out. You would have to use $C_{V1} = \frac{3}{2}R$ and $C_{V2} = \frac{5}{2}R$ in the equation.

Q: Why is the final temperature not just the simple average of 200 K and 400 K?

Because the amounts (moles) of the two gases are different. You have twice as much of the colder gas ($0.1$ mol) compared to the hotter gas ($0.05$ mol). Therefore, the final temperature will be closer to the colder gas's temperature. It's a weighted average.

Q: Is it necessary to convert temperatures to Kelvin?

In this specific problem, they are already given in Kelvin! But as a golden rule, always use absolute temperature (Kelvin) in thermodynamics formulas like $U = n C_V T$ to avoid catastrophic errors.

Question

A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to .......

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Gases

Imagine you have a sturdy, closed vessel. Inside this vessel, you have a sample of a monatomic ideal gas—let's say it's Helium or Neon. You have exactly $0.1$ moles of this gas, and it's sitting at a chilly $200	ext{ K}$.

Now, you decide to pump more of the exact same gas into this vessel. You add $0.05$ moles, but this new batch is much hotter, at $400	ext{ K}$. The question is: once these two batches of gas mix and settle down, what will be their final equilibrium temperature?

### The Master Principle: Conservation of Energy

Whenever you are dealing with a closed, insulated system where no heat is lost to the surroundings and no mechanical work is done, you have a powerful tool at your disposal: **The Conservation of Internal Energy**.

The total internal energy of the system before mixing must perfectly equal the total internal energy after mixing. We can write this mathematically as:

$$U_1 + U_2 = U_{	ext{final}}$$

For an ideal gas, the internal energy $U$ is purely a function of its temperature and is given by the formula $U = n C_V T$, where $n$ is the number of moles and $C_V$ is the molar heat capacity at constant volume.

### The Mathematical Symphony

Let's substitute our internal energy formula into the conservation equation. Since both batches are the exact same monatomic gas, their $C_V$ values are identical.

$$n_1 C_V T_1 + n_2 C_V T_2 = (n_1 + n_2) C_V T$$

Do you notice something beautiful here? The term $C_V$ appears in every single part of the equation. Because it's a non-zero constant, we can divide the entire equation by $C_V$, completely eliminating it from our math!

$$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T$$

This simplified equation tells us that the final temperature is essentially a weighted average of the initial temperatures, weighted by the number of moles.

### The Final Calculation

Now, it's just a matter of carefully plugging in our given values. We have $n_1 = 0.1	ext{ mol}$, $T_1 = 200	ext{ K}$, $n_2 = 0.05	ext{ mol}$, and $T_2 = 400	ext{ K}$.

$$(0.1)(200) + (0.05)(400) = (0.1 + 0.05) T$$

Let's compute the terms on the left side:

$$20 + 20 = 0.15 T$$

$$40 = 0.15 T$$

To find the final temperature $T$, we simply divide $40$ by $0.15$:

$$T = \frac{40}{0.15} = 266.66... 	ext{ K}$$

The problem asks for the answer rounded to the nearest integer. Rounding $266.66$ gives us our final answer:

**$$T \approx 266 	ext{ K}$$**

And there you have it! By trusting the conservation of energy, we easily found the equilibrium state of the mixture.

Animated Solution for Physics - Thermodynamics: A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to .......

Enter Numerical Value:

Visualized Solution

Visualizing the Mixing Process

Conservation of Internal Energy

Substituting the Energy Formula

Simplifying the Equation

Plugging in the Values

Calculating the Terms

Final Temperature

What if the gases were different?

The Sigma Insight: First Law of Thermodynamics

The Setup

The Master Principle

The Mathematical Symphony

The Final Calculation

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