A cylindrical container of volume 4.0 x 10^-3 m^3 contains one mole of hydrogen and two moles of car

Q: Why can we add the moles of different gases together?

According to Dalton's Law of Partial Pressures, in a mixture of non-reacting ideal gases, each gas behaves independently. The total pressure is the sum of their partial pressures, which mathematically simplifies to treating the total moles as the sum of individual moles in the ideal gas equation.

Q: Does the size or mass of the gas molecules affect the pressure?

No, for an ideal gas, the pressure depends only on the number of moles, temperature, and volume. The identity or mass of the individual gas molecules does not affect the pressure.

Q: What if the gases react with each other?

If the gases react, the total number of moles might change, and the ideal gas assumption for the initial mixture would no longer hold. You would need to find the final number of moles after the reaction reaches equilibrium.

Q: Why did we convert $2490 \times 10^3$ to $24.9 \times 10^5$?

This was done simply to match the format of the options provided in the question. Both values represent the exact same physical pressure.

Q: Can we use a different value for the gas constant $R$?

The value of $R$ must match the units of the other variables. Since pressure is in Pascals (SI unit), volume in $m^3$, and temperature in Kelvin, we must use the SI value of $R$, which is $8.3 \text{ J mol}^{-1}\text{K}^{-1}$.

Question

A cylindrical container of volume $4.0 	imes 10^{-3} 	ext{ m}^3$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is $400 	ext{ K}$. The pressure of the mixture of gases is [Take, gas constant $= 8.3 	ext{ J mol}^{-1}	ext{K}^{-1}$]

Options:

(A) $249 	imes 10^1 	ext{ Pa}$

(B) $24.9 	imes 10^3 	ext{ Pa}$

(C) $24.9 	imes 10^5 	ext{ Pa}$

(D) $24.9 	ext{ Pa}$

Sigma Percentile · Accepted Answer

### Visualizing the Gas Mixture

Imagine you are looking at a sealed cylindrical container. Inside this container, a lively dance of molecules is taking place. We have a mixture of two different gases: **one mole of hydrogen** and **two moles of carbon dioxide**.

The container has a fixed volume of $4.0 	imes 10^{-3} 	ext{ m}^3$, and the entire system is kept at a warm temperature of $400 	ext{ K}$.

Our goal is to figure out the total pressure exerted by these bouncing molecules on the walls of the container.

### The Master Equation

When dealing with a mixture of non-reacting gases, we can rely on a beautiful principle known as **Dalton's Law of Partial Pressures**.

This law tells us that each gas acts as if it were alone in the container. The total pressure is simply the sum of the pressures each gas would exert individually.

Mathematically, this means we can treat the entire mixture as one single ideal gas. We just need to find the total number of moles!

So, the total moles $n$ is the sum of the moles of hydrogen and carbon dioxide:
$$n = 1 + 2 = 3 	ext{ moles}$$

Now, we bring in our master tool, the **Ideal Gas Equation**:
$$pV = nRT$$

We want to find the pressure $p$, so let's rearrange the equation:
$$p = \frac{nRT}{V}$$

### Crunching the Numbers

With our equation ready, it's time to substitute the known values. We have $n = 3 	ext{ moles}$, the gas constant $R = 8.3 	ext{ J mol}^{-1}	ext{K}^{-1}$, the temperature $T = 400 	ext{ K}$, and the volume $V = 4.0 	imes 10^{-3} 	ext{ m}^3$.

Let's plug them in:
$$p = \frac{3 	imes 8.3 	imes 400}{4.0 	imes 10^{-3}}$$

First, let's compute the numerator. Multiplying the moles, the gas constant, and the temperature:
$$3 	imes 8.3 	imes 400 = 9960$$

Now, our expression looks much simpler:
$$p = \frac{9960}{4.0 	imes 10^{-3}}$$

### The Final Reveal

To finish the calculation, we divide $9960$ by $4.0 	imes 10^{-3}$.

Bringing the $10^{-3}$ from the denominator to the numerator changes its sign, making it $10^3$.

Now, we just divide $9960$ by $4$:
$$p = 2490 	imes 10^3 	ext{ Pa}$$

To match the options given in the question, we can rewrite this in standard scientific notation. By shifting the decimal point two places to the left, we increase the power of ten by two:
$$p = 24.9 	imes 10^5 	ext{ Pa}$$

And there we have it! The total pressure of the gas mixture is **$24.9 	imes 10^5 	ext{ Pa}$**, which perfectly matches option (c).

Select Answer:

Visualized Solution

$V = 4.0 \times 1 0^{- 3} m^{3}, T = 400 K$

$n = n_{1} + n_{2}$

$p = \frac{n R T}{V}$

$p = \frac{3 \times 8.3 \times 400}{4.0 \times 1 0 ^{- 3}}$

$p = \frac{9960}{4.0 \times 1 0 ^{- 3}}$

$p = 24.9 \times 1 0^{5} Pa$

$Conclusion$

The Sigma Insight: Kinetic Theory of Gases

Visualizing the Gas Mixture

The Master Equation

Crunching the Numbers

The Final Reveal

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=4.0×10−3 m3,T=400 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=n1​+n2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=VnRT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=4.0×10−33×8.3×400​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=4.0×10−39960​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=24.9×105 Pa

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