The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square v

Q: Why do we use Kelvin instead of Celsius?

The formula for RMS velocity is derived from the kinetic theory of gases, which relies on absolute temperature. At $0\text{ K}$, molecular motion theoretically stops. Celsius does not start at absolute zero, so the proportionality $v_{\text{rms}} \propto \sqrt{T}$ would not hold.

Q: Does this proportionality apply to all gases?

Yes, as long as the gas behaves ideally and does not undergo any chemical reaction or dissociation. If the gas molecules were to dissociate (e.g., $O_2$ breaking into $O$ atoms), the molar mass $M$ would change, and we would have to account for that.

Q: What happens to the average kinetic energy of the gas?

Unlike velocity, the average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature ($E_k = \frac{3}{2}kT$). So, if the temperature increases by a factor of 4, the kinetic energy also increases by a factor of 4.

Q: Why is the velocity proportional to $\sqrt{T}$ and not $T$?

Kinetic energy is proportional to temperature ($E_k \propto T$). Since kinetic energy is given by $\frac{1}{2}mv^2$, we have $v^2 \propto T$. Taking the square root of both sides gives us $v \propto \sqrt{T}$.

Q: What if the gas was changed to a different gas at the same time?

If the gas is changed, the molar mass $M$ changes. You would then use the full proportionality $v_{\text{rms}} \propto \sqrt{\frac{T}{M}}$ and set up the ratio as $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1} \cdot \frac{M_1}{M_2}}$.

Question

The temperature of an ideal gas is increased from $120	ext{ K}$ to $480	ext{ K}$. If at $120	ext{ K}$ the root mean square velocity of the gas molecules is $v$, at $480	ext{ K}$ it becomes

Options:

(A) $4 v$

(B) $2 v$

(C) $v/2$

(D) $v/4$

Sigma Percentile · Accepted Answer

Welcome to a fascinating journey into the microscopic world of gases! Have you ever wondered what happens to the tiny, invisible molecules of a gas when you turn up the heat? This problem gives us a front-row seat to that exact phenomenon. We are going to explore the beautiful relationship between the temperature of a gas and the speed at which its molecules zip around.

I know that dealing with formulas and square roots can sometimes feel a bit abstract, but let's take a breath and visualize the physical reality. Imagine you are looking at a sealed container filled with an ideal gas. The molecules are constantly moving, colliding with each other and the walls. This chaotic dance is what we measure as temperature!

### Analyzing the Setup

In our problem, we are given an ideal gas initially sitting at a temperature of $120	ext{ K}$. At this state, the root mean square (RMS) velocity of its molecules is $v$.

Then, we crank up the heat! We increase the temperature all the way to $480	ext{ K}$. Our mission is to find out what happens to the RMS velocity of these molecules at this new, hotter state.

To solve this, we need a mathematical tool that connects temperature and velocity. Enter the Kinetic Theory of Gases!

### The Master Equation

The cornerstone of our solution is the formula for the root mean square velocity of an ideal gas:

$$v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$$

Let's break this down. $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $M$ is the molar mass of the gas.

Notice something crucial here: we are dealing with the **same** gas throughout the entire process. This means that the molar mass $M$ does not change. And, of course, $3$ and $R$ are constants.

When we strip away all the constants, we are left with a beautiful, elegant proportionality:

**$v_{	ext{rms}} \propto \sqrt{T}$**

This tells us that the speed of the molecules scales with the square root of the absolute temperature. It’s a profound insight! If you want to double the speed of the molecules, you can't just double the temperature; you have to quadruple it!

### Setting Up the Ratio

Since we are comparing two different states of the same gas, the most efficient way to solve this is by setting up a ratio. We can write:

$$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$

This equation is our bridge from the initial state to the final state. It perfectly captures the proportionality we just discovered.

### Final Calculation

Now, let's plug in the numbers from our problem. We know our initial temperature $T_1$ is $120	ext{ K}$, and our final temperature $T_2$ is $480	ext{ K}$. The initial velocity $v_1$ is simply $v$.

$$\frac{v_2}{v} = \sqrt{\frac{480}{120}}$$

Let's simplify the fraction inside the square root. $480$ divided by $120$ is exactly $4$.

$$\frac{v_2}{v} = \sqrt{4}$$

And we all know that the square root of $4$ is $2$.

$$\frac{v_2}{v} = 2$$

Finally, we multiply both sides by $v$ to isolate our final velocity:

**$v_2 = 2v$**

And there we have it! By increasing the temperature by a factor of four (from $120	ext{ K}$ to $480	ext{ K}$), the root mean square velocity of the gas molecules has exactly doubled.

This is a classic, high-yield concept for competitive exams. Always remember to check whether the question is asking about velocity or kinetic energy. While velocity scales with the square root of temperature, kinetic energy scales directly with temperature. Keep this distinction clear, and you'll master these problems every single time!

Animated Solution for Physics - Thermodynamics: The temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ the root mean square velocity of the gas molecules is $v$ , at $480 K$ it becomes

Select Answer:

Visualized Solution

Visualizing the Two States

Formula for RMS Velocity

Establishing Proportionality

Setting up the Ratio

Substituting the Values

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation

Setting Up the Ratio

Final Calculation

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