Number of molecules in a volume of 4 cm^3 of a perfect monoatomic gas at some temperature T and at a

Q: Why do we use $U = \frac{3}{2} k_B T$?

For a perfect monoatomic gas, the only form of internal energy is translational kinetic energy. A monoatomic molecule has 3 degrees of freedom, and by the equipartition theorem, each degree contributes $\frac{1}{2} k_B T$.

Q: Do I need to convert the units to SI (meters, kilograms, Joules)?

Not necessarily! In this problem, every single value is given in the CGS system (centimeters, grams, seconds, ergs). If you stick to CGS throughout, your final answer for the number of molecules (which is dimensionless) will be perfectly correct.

Q: What does the '2 cm of mercury' mean?

It's a way to measure pressure. It means the gas exerts the same pressure as the weight of a 2 cm tall column of mercury spread over a given area. We calculate this using the hydrostatic pressure formula $p = \rho g h$.

Q: Why did we substitute $k_B T$ instead of finding $T$ first?

Finding $T$ first would require knowing the value of the Boltzmann constant $k_B$. By substituting $k_B T = \frac{2}{3} U$ directly into the ideal gas law, $k_B$ cancels out, saving us a step and avoiding the need to memorize its value!

Q: Why is the answer approximate?

The exact calculation gives $3.9984 \times 10^{18}$. In physics, we often round to the appropriate number of significant figures based on the given data. Here, $4.0 \times 10^{18}$ is the closest and most reasonable approximation among the options.

Question

Number of molecules in a volume of $4 	ext{ cm}^3$ of a perfect monoatomic gas at some temperature $T$ and at a pressure of $2 	ext{ cm}$ of mercury is close to
(Given, mean kinetic energy of a molecule at $T$ is $4 	imes 10^{-14} 	ext{ erg}$, $g = 980 	ext{ cm/s}^2$, density of mercury $= 13.6 	ext{ g/cm}^3$)

Options:

(A) $5.8 	imes 10^{16}$

(B) $4.0 	imes 10^{16}$

(C) $4.0 	imes 10^{18}$

(D) $5.8 	imes 10^{18}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine a small container holding a perfect monoatomic gas. We are given that its volume is $V = 4 	ext{ cm}^3$. The pressure it exerts is not given directly in Pascals or atmospheres, but rather as the equivalent of a $2 	ext{ cm}$ column of mercury. This is a classic way to represent pressure using a manometer setup.

### The Ideal Gas Law

To find the number of molecules, we need an equation that links our known macroscopic variables (pressure and volume) to the microscopic count of molecules. The ideal gas equation in terms of the Boltzmann constant $k_B$ is perfect for this:

$$pV = N k_B T$$

Rearranging this to solve for the number of molecules $N$, we get:

$$N = \frac{pV}{k_B T}$$

### Kinetic Energy Connection

We have a slight problem: we don't know the absolute temperature $T$. However, the problem provides the mean kinetic energy $U$ of a single monoatomic molecule. According to the equipartition of energy, a monoatomic gas molecule has 3 translational degrees of freedom, so its mean kinetic energy is:

$$U = \frac{3}{2} k_B T$$

We can rearrange this to isolate the $k_B T$ term:

$$k_B T = \frac{2}{3} U$$

### Bringing It All Together

Now, let's substitute this expression for $k_B T$ back into our equation for $N$:

$$N = \frac{pV}{\frac{2}{3} U} = \frac{3pV}{2U}$$

We also need to express the pressure $p$ in terms of the mercury column. The hydrostatic pressure formula is $p = ho g h$. Substituting this into our equation gives us our master formula:

$$N = \frac{3 ho g h V}{2U}$$

### The Final Calculation

Here is where we must be careful with units. Notice that every single value given in the problem is in the CGS (centimeter-gram-second) system:
- $ho = 13.6 	ext{ g/cm}^3$
- $g = 980 	ext{ cm/s}^2$
- $h = 2 	ext{ cm}$
- $V = 4 	ext{ cm}^3$
- $U = 4 	imes 10^{-14} 	ext{ erg}$

Because everything is consistently in CGS, we can substitute the numbers directly without any messy conversions!

$$N = \frac{3 	imes 13.6 	imes 980 	imes 2 	imes 4}{2 	imes 4 	imes 10^{-14}}$$

The $2$ and $4$ in the numerator and denominator cancel out beautifully, leaving:

$$N = \frac{3 	imes 13.6 	imes 980}{10^{-14}}$$

$$N = 39984 	imes 10^{14}$$

Converting this to standard scientific notation, we get:

$$N = 3.9984 	imes 10^{18}$$

This is approximately **$4.0 	imes 10^{18}$** molecules, which perfectly matches option (c).

Select Answer:

Visualized Solution

Visualizing the Setup

The Ideal Gas Equation

Kinetic Energy of Monoatomic Gas

Substituting Temperature

Substituting Pressure

Plugging in the Values

Simplifying the Expression

Final Answer

The Sigma Insight: Kinetic Theory of Gases

Visualizing the Setup

The Ideal Gas Law

Kinetic Energy Connection

Bringing It All Together

The Final Calculation

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