The molecules of a given mass of a gas have root mean square speeds of $100\ \text{m s}^{-1}$ at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

The molecules of a given mass of a gas have root mean square speeds of 100m s⁻¹ at 27°C and 1.00 atmospheric p

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100\ \text{m s}^{-1}

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Concept understanding — Kinetic Theory of Gases

Kinetic Theory of Gases

Imagine you're sitting in a quiet room. The air around you feels still — but it isn't. Every second, billions of tiny particles (molecules of nitrogen, oxygen, and others) are zipping past you at hundreds of metres per second. They're constantly crashing into each other and into the walls, your skin, the furniture. You don't feel each individual hit because the molecules are so small and the collisions happen so fast. But collectively, those countless tiny impacts produce something you do feel: pressure.

That's the core intuition behind the kinetic theory of gases. It says: all the macroscopic properties of a gas — pressure, temperature, volume — can be explained by the motion of its molecules.

The Big Idea

Instead of treating a gas as a continuous, smooth substance (like a fluid), the kinetic theory treats it as a swarm of tiny, hard, perfectly elastic balls in constant, random motion. "Perfectly elastic" means that when two molecules collide, no kinetic energy is lost — they bounce off each other like ideal billiard balls, not like sticky clay.

From this simple picture, we can derive the gas laws (Boyle's, Charles's, Avogadro's) and even calculate things like the speed of sound in a gas.

The Five Assumptions (The Precise Statement)

For a gas to behave according to the kinetic theory in its simplest form, we make these assumptions:

A gas consists of a very large number of molecules.

The number is so huge that we can use statistics — individual molecules don't matter, only averages do.
The molecules are in constant, random motion.

They move in straight lines until they hit something (another molecule or a wall). There's no preferred direction.
The molecules are point masses.

Their actual size is negligible compared to the distance between them. In other words, the volume of the molecules themselves is tiny compared to the volume of the container.
Collisions are perfectly elastic.

No kinetic energy is lost when molecules collide with each other or with the walls. Total energy of the system stays constant.
There are no intermolecular forces.

The molecules don't attract or repel each other except during collisions. Between collisions, they move freely.

Important

These assumptions define an ideal gas. Real gases deviate from this behaviour at high pressure or low temperature, but the kinetic theory gives an excellent approximation for most everyday conditions.

How It Explains Pressure

Pressure is the force per unit area exerted by the gas on the walls of its container. In the kinetic picture:

A molecule moving toward a wall hits it and bounces back.
During the collision, the wall exerts a force on the molecule to reverse its momentum.
By Newton's third law, the molecule exerts an equal and opposite force on the wall.
Multiply that by the billions of collisions happening every second, and you get a steady, measurable pressure.

If you heat the gas, the molecules move faster. They hit the walls harder and more often — pressure increases. If you compress the gas into a smaller volume, molecules hit the walls more frequently — pressure increases again.

The Key Result: The Kinetic Equation

From these assumptions, we can derive a relationship between pressure $P$ , volume $V$ , and the average kinetic energy of the molecules. The result is:

P V = \frac{1}{3} N m \overline{v^2}

Where:

$N$ = number of molecules
$m$ = mass of one molecule
$\overline{v^2}$ = mean square speed of the molecules (average of the squares of their speeds)

Since the average kinetic energy of a molecule is $\overline{K} = \frac{1}{2} m \overline{v^2}$ , we can rewrite this as:

$P V = \frac{2}{3} N \overline{K}$ …

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The root mean square speed depends only on temperature, not pressure. Raising the temperature from 27°C to 127°C increases $v_{\text{rms}}$ by a factor of $\sqrt{400/300}$ , giving 115.5 m s⁻¹.

The root mean square speed is a measure of the typical speed of gas molecules arising from their thermal energy. The kinetic theory of gases tells us that the average translational kinetic energy of a molecule is directly proportional to absolute temperature: $\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_BT$ , where $k_B$ is Boltzmann's constant. This immediately reveals that $v_{\text{rms}} = \sqrt{\langle v^2\rangle}$ depends on temperature but is independent of pressure.

v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}

where $m$ is the molecular mass, $M$ is the molar mass, and $R$ is the gas constant.

The pressure appears nowhere in this expression. Pressure affects the number density of molecules (how many collisions occur per unit area per unit time), but each molecule's speed distribution is governed solely by temperature.

Solution

Convert temperatures to Kelvin. The absolute temperature scale is required because kinetic energy is proportional to $T$ , not Celsius temperature.

T_1 = 27 + 273 = 300\ \text{K}

T_2 = 127 + 273 = 400\ \text{K}

Recognize that pressure is irrelevant.

The change from 1.00 atm to 2.0 atm does not affect molecular speeds. The gas molecules move faster or slower based only on how much thermal energy they possess, which is set by temperature.
Apply the temperature dependence of $v_{\text{rms}}$ .

Since $v_{\text{rms}} \propto \sqrt{T}$ , we can write:

$\frac{v_{\text{rms},2}}{v_{\text{rms},1}} = \sqrt{\frac{T_2}{T_1}}$ …

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Q.The molecules of a given mass of a gas have root mean square speeds of 100 m s−1100\ \text{m s}^{-1}100 m s−1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

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Q.The molecules of a given mass of a gas have root mean square speeds of $100\ \text{m s}^{-1}$ at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

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