The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and

Q: Why don't we consider the volume of individual gases?

In the kinetic theory of gases, ideal gas molecules are considered point masses that don't interact. Therefore, every single gas in the mixture occupies the entire volume $V$ of the container independently.

Q: What if the gases reacted with each other?

If the gases reacted, the total number of moles would change according to the stoichiometry of the reaction. However, at normal temperatures, oxygen, nitrogen, and carbon dioxide do not react, so we simply add their moles.

Q: Why use $5/2$ instead of $2.5$?

While $2.5$ is mathematically identical, JEE options frequently use fractional forms like $5/2$. It's a good habit to keep your intermediate steps in fractions to easily match the final options.

Q: Does the identity of the gas matter for the pressure it exerts?

Fascinatingly, no! For an ideal gas, the pressure depends only on the *number* of molecules (or moles), not their mass or chemical identity. A mole of heavy $CO_2$ exerts the exact same pressure as a mole of light $H_2$.

Q: Can I calculate partial pressures first and then add them?

Absolutely! That is Dalton's Law of Partial Pressures ($p_{\text{total}} = p_1 + p_2 + p_3$). Since $p_i = n_i RT/V$, summing the partial pressures is mathematically identical to summing the moles first.

Question

The volume $V$ of an enclosure contains a mixture of three gases, $16 	ext{ g}$ of oxygen, $28 	ext{ g}$ of nitrogen and $44 	ext{ g}$ of carbon dioxide at absolute temperature $T$. Consider $R$ as universal gas constant. The pressure of the mixture of gases is

Options:

(A) $\frac{88 RT}{V}$

(B) $\frac{3RT}{V}$

(C) $\frac{5}{2}\frac{RT}{V}$

(D) $\frac{4RT}{V}$

Sigma Percentile · Accepted Answer

Imagine you are looking at a sealed container. Inside, a chaotic but beautiful dance is taking place.

We have a mixture of three different gases: oxygen, nitrogen, and carbon dioxide. They are all trapped together in a volume $V$ and held at a constant absolute temperature $T$.

Our mission is to find the total pressure exerted by this molecular party on the walls of the container.

### The Master Equation

When dealing with a mixture of non-reacting ideal gases, we have a powerful tool at our disposal. We can treat the entire mixture as if it were just one single ideal gas!

This means the classic ideal gas equation still applies perfectly.

$$pV = n_{	ext{total}}RT$$

Here, **the total pressure depends only on the total number of moles of gas present**, regardless of what types of gases they are.

To find the pressure $p$, we just need to count the total number of moles $n_{	ext{total}}$ and rearrange our equation.

### Counting the Guests

Let's calculate the number of moles for each gas individually. We do this by dividing the given mass by the molar mass of the gas.

First, let's look at oxygen ($O_2$). We are given $16 	ext{ g}$ of oxygen. Knowing that the molar mass of $O_2$ is $32 	ext{ g/mol}$, we can find its moles.

$$n_{O_2} = \frac{16}{32} = \frac{1}{2} 	ext{ mol}$$

Next up is nitrogen ($N_2$). We have $28 	ext{ g}$ of nitrogen gas. The molar mass of $N_2$ is exactly $28 	ext{ g/mol}$.

$$n_{N_2} = \frac{28}{28} = 1 	ext{ mol}$$

Finally, we have carbon dioxide ($CO_2$). The problem gives us $44 	ext{ g}$ of it. Conveniently, the molar mass of $CO_2$ is also $44 	ext{ g/mol}$.

$$n_{CO_2} = \frac{44}{44} = 1 	ext{ mol}$$

### The Grand Finale

Now that we have the moles of each individual gas, we simply add them up to find the total number of moles in the container.

$$n_{	ext{total}} = \frac{1}{2} + 1 + 1 = \frac{5}{2} 	ext{ mol}$$

We are ready for the final step. We substitute this total number of moles back into our rearranged ideal gas equation.

$$p = \frac{n_{	ext{total}}RT}{V}$$

Plugging in our value for $n_{	ext{total}}$, we arrive at our final answer.

$$p = \frac{5}{2}\frac{RT}{V}$$

**This is the total pressure exerted by the mixture of gases.**

Notice how elegantly Dalton's Law of Partial Pressures aligns with this. If we had calculated the pressure of each gas separately and added them together, we would have reached the exact same beautiful result!

Animated Solution for Physics - Thermodynamics: The volume $V$ of an enclosure contains a mixture of three gases, $16 g$ of oxygen, $28 g$ of nitrogen and $44 g$ of carbon dioxide at absolute temperature $T$ . Consider $R$ as universal gas constant. The pressure of the mixture of gases is

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Visualized Solution

$The Gas Mixture$

$Ideal Gas Equation for Mixture$

$Moles of Oxygen (O_{2})$

$Moles of Nitrogen (N_{2})$

$Moles of Carbon Dioxide (C O_{2})$

$Total Moles (n_{total})$

$Final Pressure$

$Dalton’s Law Perspective$

The Sigma Insight: Kinetic Theory of Gases

The Master Equation

Counting the Guests

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Gas Mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Equation for Mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Oxygen (O2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Nitrogen (N2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Carbon Dioxide (CO2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Moles (ntotal​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Pressure

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dalton’s Law Perspective