Part 2 · Chapter 19

The Kinetic Theory of Gases

How the unseen storm of molecules — each faster than a bullet, colliding billions of times a second — adds up to the pressure, temperature, and warmth of an ordinary breath of air

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 65 min

i What you'll learn

How to count atoms by the mole using Avogadro's number \(N_{A} = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}\), and the ideal gas law in its two forms \(pV = nRT = NkT\).
The work an ideal gas does — isothermal \(W = nRT\ln(V_{f}/V_{i})\), constant-pressure \(W = p\,\Delta V\), constant-volume \(W = 0\).
That pressure springs from molecular motion, \(p = nMv_{\text{rms}}^{2}/3V\), giving the rms speed \(v_{\text{rms}} = \sqrt{3RT/M}\).
That temperature is average translational kinetic energy: \(K_{\text{avg}} = \tfrac{3}{2}kT\).
The mean free path \(\lambda = 1/(\sqrt{2}\,\pi d^{2}\,N/V)\) and the Maxwell speed distribution with its three speeds \(v_{P} < v_{\text{avg}} < v_{\text{rms}}\).
The molar specific heats \(C_{V} = \tfrac{f}{2}R\), \(C_{p} = C_{V} + R\), the role of degrees of freedom, and the adiabatic law \(pV^{\gamma} = \text{const}\).

Section 19-1

What Is Physics?

A gas is a swarm of atoms or molecules that fill their container and press on its walls. Three everyday quantities describe it — volume, pressure, and temperature — and all three trace back to the same hidden cause: the motion of the molecules. The volume reflects their freedom to roam, the pressure their relentless drumming on the walls, and the temperature their kinetic energy. The kinetic theory of gases is the bridge that connects this unseen molecular motion to the measurable, large-scale behavior of the gas.

Its applications are everywhere: the combustion of vaporized fuel in an engine, the fermentation gas that lifts a loaf of bread, the cork launched from a champagne bottle, the nitrogen a diver must let bleed slowly from the blood to avoid the bends, and the heat traded between ocean and air that shapes weather. Our first step is simply to count: how much gas is in a sample? For that, we use Avogadro's number.

Section 19-2

Avogadro's Number

When we think about atoms, it is natural to measure samples in moles, so that equal counts of molecules are being compared. One mole is the number of atoms in a 12 g sample of carbon-12, and that number — determined experimentally — is Avogadro's number:

Avogadro's number, moles, and molar mass

\[ N_{A} = 6.02 \times 10^{23}\,\mathrm{mol^{-1}} \qquad n = \frac{N}{N_{A}} = \frac{M_{\text{sam}}}{M} = \frac{M_{\text{sam}}}{mN_{A}} \qquad M = mN_{A} \]

Sort the symbols now to avoid "N-confusion": \(N\) is the number of molecules, \(n\) the number of moles, \(N_A\) molecules per mole. The molar mass \(M\) (mass of one mole) equals the molecular mass \(m\) times \(N_A\).

Avogadro's insight. Amedeo Avogadro proposed that equal volumes of any gases, at the same temperature and pressure, hold the same number of molecules. That single idea — that what matters for a gas is the count of particles, not their identity — is the seed of everything in this chapter.

Section 19-3

Ideal Gases

We want to explain a gas's large-scale behavior from its molecules — but which gas? Hydrogen, oxygen, methane, and uranium hexafluoride are all different. Yet experiment reveals something wonderful: confine one mole of any gas in a fixed volume at a fixed temperature and the measured pressures are nearly the same, and the small differences vanish as the density drops. At low enough density, every real gas obeys one law:

⚗️

The Ideal Gas Law

pV = nRT = NkT

Here \(p\) is absolute pressure, \(T\) is in kelvins, \(R = 8.31\,\mathrm{J/mol\cdot K}\) is the gas constant, and \(k = R/N_{A} = 1.38 \times 10^{-23}\,\mathrm{J/K}\) is the Boltzmann constant. The first form counts moles (\(n\)); the second counts molecules (\(N\)). An "ideal" gas is the limit in which molecules are so far apart that they ignore one another — and all real gases approach it at low density.

What crushed the tank car. A sealed railroad tank car, cleaned with hot steam and then closed up, was found crushed the next morning. As it cooled overnight, steam condensed, so both \(N\) and \(T\) in \(pV = NkT\) fell. With \(V\) fixed, the internal pressure dropped until the outside atmosphere simply caved the steel walls in. Leaving a valve open would have prevented it.

Section 19-4

Work Done by an Ideal Gas

Put an ideal gas in a piston-cylinder and let it expand at constant temperature — an isothermal process, traced on a p–V diagram by a curve called an isotherm. Starting from the general \(W = \int p\,dV\) and substituting \(p = nRT/V\), the constant \(T\) comes out of the integral:

Work in three standard processes

\[ W_{\text{isothermal}} = nRT\ln\frac{V_{f}}{V_{i}} \qquad W_{\text{const-}V} = 0 \qquad W_{\text{const-}p} = p\,\Delta V \]

For an expansion \(V_f > V_i\), the logarithm is positive, so the gas does positive work — as expected. At constant volume the gas cannot push anything, so \(W = 0\). At constant pressure \(p\) slides outside the integral, leaving \(p\,\Delta V\).

Read it off the diagram. Work is always the area under the path on a p–V diagram. Because different paths between the same two states enclose different areas, \(W\) (and the heat \(Q\)) depend on the route — exactly the path-dependence introduced with the first law in Chapter 18.

Section 19-5

Pressure, Temperature, and RMS Speed

Now the first true kinetic-theory result. Picture \(n\) moles in a cubical box, molecules ricocheting off the walls like balls in a court. Each elastic bounce off a wall reverses one velocity component and delivers momentum \(2mv_{x}\); the molecule returns every \(2L/v_{x}\) seconds. Sum the resulting force over all molecules, divide by the wall area, and the macroscopic pressure emerges from microscopic speeds:

Pressure from molecular motion, and the rms speed

\[ p = \frac{nM\,v_{\text{rms}}^{2}}{3V} \qquad\qquad v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]

The root-mean-square speed is exactly what its name says: square every speed, average, then take the square root. Combining the pressure result with \(pV = nRT\) gives \(v_{\text{rms}}\) directly from temperature and molar mass.

Table 19-1 · Some rms speeds at 300 K
Gas	M (10⁻³ kg/mol)	v_rms (m/s)
Hydrogen (H₂)	2.02	1920
Helium (He)	4.0	1370
Water vapor (H₂O)	18.0	645
Nitrogen (N₂)	28.0	517
Oxygen (O₂)	32.0	483
Carbon dioxide (CO₂)	44.0	412

Faster than a bullet — so why slow perfume? Hydrogen molecules at room temperature average 1920 m/s, over 4000 mi/h. Yet it takes a minute to smell perfume opened across the room. The resolution is the mean free path (next): each molecule's blistering speed is squandered on billions of collisions per second, so it diffuses across the room only slowly. Notice too that lighter molecules move faster — the heavier the gas, the lower its \(v_{\text{rms}}\).

Section 19-6

Translational Kinetic Energy

Follow a single molecule as collisions keep changing its speed. Its average translational kinetic energy is \(K_{\text{avg}} = \tfrac{1}{2}mv_{\text{rms}}^{2}\). Substitute \(v_{\text{rms}} = \sqrt{3RT/M}\), recognize that \(M/m = N_{A}\), and the molar mass and molecular mass cancel beautifully to leave something universal:

🌡️

Temperature is average kinetic energy

K_avg = (3/2) kT

At a given temperature, every ideal-gas molecule has the same average translational kinetic energy — regardless of its mass. A heavy molecule simply moves more slowly to compensate. So when you measure a gas's temperature, you are really measuring the average translational kinetic energy of its molecules. This is the deep meaning of temperature.

Section 19-7

Mean Free Path

A molecule does not fly straight across the room; it zigzags, colliding constantly. The mean free path \(\lambda\) is the average distance it travels between collisions. It should shrink when the gas is crowded (large \(N/V\)) and when molecules are fat (large diameter \(d\)) — and because what matters is a molecule's cross-sectional target area, it falls off as \(d^{2}\):

Mean free path

\[ \lambda = \frac{1}{\sqrt{2}\,\pi d^{2}\,N/V} \]

A molecule sweeps a cylinder of cross-section \(\pi d^2\); the number of point-molecules inside it counts the collisions. The factor \(\sqrt{2}\) appears once we account for the fact that the other molecules are moving too, with \(v_{\text{rel}} = \sqrt{2}\,v_{\text{avg}}\).

From a fraction of a millimetre to kilometres. For air at sea level \(\lambda \approx 0.1\,\mu\mathrm{m}\). At 100 km altitude the air is so thin that \(\lambda\) grows to about 16 cm; at 300 km it reaches roughly 20 km. That is why upper-atmosphere chemistry is so hard to reproduce in a lab — no container is large enough to mimic the emptiness.

Section 19-8

The Distribution of Molecular Speeds

The rms speed is a single summary, but molecules actually carry a whole spread of speeds. In 1852 James Clerk Maxwell found the distribution. The function \(P(v)\) is a probability density: \(P(v)\,dv\) is the fraction of molecules with speeds in a band \(dv\) around \(v\), and the total area under the curve is 1.

Maxwell speed distribution, and three characteristic speeds

\[ P(v) = 4\pi\!\left(\frac{M}{2\pi RT}\right)^{3/2}\!v^{2}\,e^{-Mv^{2}/2RT} \]

\[ v_{P} = \sqrt{\frac{2RT}{M}} \;<\; v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \;<\; v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]

The most probable speed \(v_P\) sits at the peak; the average \(v_{\text{avg}}\) is a bit higher; the rms speed is highest of all, because squaring weights the fast molecules most. Heating shifts the whole curve right and flattens it.

Why it rains and why the Sun shines. Both depend on the high-speed tail of the curve. Only the rare, fast water molecules escape a pond's surface to form clouds — evaporation. And only the rare, fast protons in the Sun's core carry enough energy to overcome their mutual repulsion and fuse. Average molecules can do neither; life depends on the exceptions.

Section 19-9

The Molar Specific Heats of an Ideal Gas

For a monatomic ideal gas, the internal energy is just the translational kinetic energy of its atoms, \(E_{\text{int}} = \tfrac{3}{2}nRT\) — a function of temperature alone. From this we derive two molar specific heats: \(C_{V}\) (heat per mole per degree at constant volume) and \(C_{p}\) (at constant pressure).

Molar specific heats and internal-energy change

\[ C_{V} = \tfrac{3}{2}R = 12.5\,\mathrm{J/mol\cdot K}\;\text{(monatomic)} \qquad C_{p} = C_{V} + R \qquad \Delta E_{\text{int}} = nC_{V}\,\Delta T \]

Crucially, \(\Delta E_{\text{int}} = nC_{V}\Delta T\) holds for any process — constant volume, constant pressure, or otherwise — because internal energy depends only on temperature. \(C_p\) exceeds \(C_V\) by exactly \(R\): at constant pressure the extra heat \(nR\,\Delta T\) pays for the work the expanding gas does.

Table 19-2 · Molar specific heats at constant volume, \(C_V\;(\mathrm{J/mol\cdot K})\)
Molecule	Ideal prediction	Real example
Monatomic	\(\tfrac{3}{2}R = 12.5\)	He 12.5 · Ar 12.6
Diatomic	\(\tfrac{5}{2}R = 20.8\)	N₂ 20.7 · O₂ 20.8
Polyatomic	\(3R = 24.9\)	CO₂ 29.7

Section 19-10

Degrees of Freedom and Molar Specific Heats

The prediction \(C_{V} = \tfrac{3}{2}R\) nails monatomic gases but falls short for diatomic and polyatomic ones — because those molecules can also store energy in rotation (and, at high temperatures, vibration). Maxwell's equipartition theorem explains it: every independent way a molecule can store energy — every degree of freedom — holds, on average, \(\tfrac{1}{2}kT\) per molecule (\(\tfrac{1}{2}RT\) per mole).

🎲

Equipartition and degrees of freedom

C_V = (f/2) R = 4.16 f J/mol·K

A monatomic atom has only 3 translational degrees (\(f = 3\)). A diatomic molecule adds 2 rotational degrees (\(f = 5\)) — it cannot rotate about its own long axis. Polyatomic molecules get all 3 rotational degrees (\(f = 6\)). Each extra storage channel raises \(C_V\), matching experiment.

A hint of quantum theory. Plot \(C_{V}/R\) for hydrogen against temperature and it climbs in steps: about 1.5 below 80 K (translation only), rising to 2.5 as rotation "turns on," then toward 3.5 near 1000 K as vibration switches on. Energy storage is not continuous — modes activate only when molecules have enough energy, a foreshadowing of quantum mechanics.

Section 19-11

The Adiabatic Expansion of an Ideal Gas

In an adiabatic process no heat crosses the boundary (\(Q = 0\)) — achieved either by good insulation or by acting so fast there is no time for heat flow, as in a sound wave. Starting from the first law with \(dE_{\text{int}} = nC_{V}\,dT\) and integrating, one finds the gas tracks a steeper curve than an isotherm:

Adiabatic process for an ideal gas (γ = C_p/C_V)

\[ pV^{\gamma} = \text{constant} \qquad\qquad TV^{\gamma-1} = \text{constant} \]

Here \(\gamma = C_p/C_V\) (for a diatomic gas with rotation, \(\gamma = \tfrac{7}{2}R / \tfrac{5}{2}R = 1.40\)). Because \(Q = 0\), any work the gas does comes straight out of its internal energy: \(W = -\Delta E_{\text{int}}\), so an adiabatic expansion cools the gas.

The fog at the champagne cork. Pop a cold bottle and a little fog appears at the neck. The trapped CO₂ and water vapor expand suddenly into the air — too fast for heat to flow, so the expansion is adiabatic. The gas does work pushing back the atmosphere, its internal energy drops, the temperature falls, and the chilled water vapor condenses into a tiny cloud. A free expansion (gas rushing into a vacuum) is different: no work, no heat, so \(T_{i} = T_{f}\) and \(p_{i}V_{i} = p_{f}V_{f}\).

Worked Examples

Putting It to Work

1 Ideal gas with changing T, V, and p

Problem. A cylinder holds 12 L of oxygen at 20 °C and 15 atm. The temperature is raised to 35 °C and the volume reduced to 8.5 L. Find the final pressure (assume ideal).

Solution. Write \(pV = nRT\) for both states and divide; with \(n\) and \(R\) cancelling, only the temperatures need converting to kelvins (293 K and 308 K).

p_f = p_i (T_f V_i)/(T_i V_f)

\[ p_{f} = (15\,\mathrm{atm})\frac{(308\,\mathrm{K})(12\,\mathrm{L})}{(293\,\mathrm{K})(8.5\,\mathrm{L})} \approx 22\,\mathrm{atm} \]

Volume and pressure units cancel in the ratio, so liters and atmospheres are fine — but temperature must be absolute, because the conversion to kelvins adds a constant that does not cancel.

2 Work in an isothermal expansion

Problem. One mole of oxygen (ideal) expands isothermally at \(T = 310\,\mathrm{K}\) from 12 L to 19 L. How much work does the gas do?

Solution. Constant temperature means \(W = nRT\ln(V_{f}/V_{i})\).

W = nRT ln(V_f / V_i)

\[ W = (1)(8.31)(310)\ln\frac{19}{12} \approx 1180\,\mathrm{J} \]

Positive, as it must be for an expansion. Reverse the process — compress 19 L back to 12 L — and the gas does −1180 J, meaning an external agent must do 1180 J on it.

3 RMS speed of oxygen at room temperature

Problem. Find the rms speed of oxygen molecules (\(M = 0.0320\,\mathrm{kg/mol}\)) at 300 K.

Solution. Apply \(v_{\text{rms}} = \sqrt{3RT/M}\) directly.

v_rms = √(3RT / M)

\[ v_{\text{rms}} = \sqrt{\frac{3(8.31)(300)}{0.0320}} \approx 483\,\mathrm{m/s} \]

For comparison, the same gas has \(v_{\text{avg}} \approx 445\,\mathrm{m/s}\) and \(v_{P} \approx 395\,\mathrm{m/s}\) — neatly confirming the ordering \(v_{P} < v_{\text{avg}} < v_{\text{rms}}\).

4 Mean free path and collision frequency

Problem. For oxygen at 300 K and 1.0 atm (\(d = 290\,\mathrm{pm}\)), find (a) the mean free path and (b) the collision frequency if the average speed is 450 m/s.

Solution. Use \(N/V = p/kT\) from the ideal gas law, substitute into \(\lambda = 1/(\sqrt{2}\,\pi d^{2}\,N/V)\), then \(f = v/\lambda\).

λ = kT/(√2 π d² p), then f = v/λ

\[ \lambda = \frac{(1.38\times10^{-23})(300)}{\sqrt{2}\,\pi(2.9\times10^{-10})^{2}(1.01\times10^{5})} \approx 1.1\times10^{-7}\,\mathrm{m} \]

\[ f = \frac{v}{\lambda} = \frac{450}{1.1\times10^{-7}} \approx 4.1\times10^{9}\,\mathrm{s^{-1}} \]

The path is about 380 molecular diameters — and the molecule suffers roughly 4 billion collisions every second, with under a nanosecond of free flight between them. That is why perfume diffuses so slowly despite its speed.

5 Monatomic gas: heat, internal energy, and work

Problem. A bubble of 5.00 mol of helium (monatomic, ideal) is warmed by \(\Delta T = 20.0\,\mathrm{C^{\circ}}\) at constant pressure. Find (a) the heat \(Q\), (b) the internal-energy change, and (c) the work done.

Solution. Constant pressure uses \(Q = nC_{p}\Delta T\) with \(C_{p} = \tfrac{5}{2}R\); internal energy uses \(\Delta E_{\text{int}} = nC_{V}\Delta T\) with \(C_{V} = \tfrac{3}{2}R\); work follows from the first law.

Q = n(5/2)R ΔT ; ΔE_int = n(3/2)R ΔT ; W = Q − ΔE_int

\[\begin{aligned} Q &= (5.00)(2.5)(8.31)(20.0) \approx 2080\,\mathrm{J} \\ \Delta E_{\text{int}} &= (5.00)(1.5)(8.31)(20.0) \approx 1250\,\mathrm{J} \\ W &= Q - \Delta E_{\text{int}} \approx 831\,\mathrm{J} \end{aligned}\]

Of the ~2080 J supplied, about 1250 J raises the internal energy (here, pure translational kinetic energy) and ~831 J becomes the work of expansion. The work can also be gotten directly as \(W = nR\,\Delta T\) — the same 831 J.

6 Adiabatic vs free expansion

Problem. One mole of oxygen (diatomic, \(\gamma = 1.40\)) at 310 K and 12 L expands to 19 L. Find the final temperature if the expansion is (a) adiabatic, and (b) a free expansion (from \(p_{i} = 2.0\,\mathrm{Pa}\)).

Solution. Adiabatic uses \(T_{i}V_{i}^{\gamma-1} = T_{f}V_{f}^{\gamma-1}\); a free expansion changes nothing about the molecules' kinetic energy, so \(T_{f} = T_{i}\).

Adiabatic: T_f = T_i (V_i/V_f)^(γ−1) ; Free: T_f = T_i

\[ \text{(a)}\quad T_{f} = (310)\!\left(\tfrac{12}{19}\right)^{0.40} \approx 258\,\mathrm{K} \]

\[ \text{(b)}\quad T_{f} = 310\,\mathrm{K}, \qquad p_{f} = p_{i}\frac{V_{i}}{V_{f}} = (2.0)\tfrac{12}{19} \approx 1.3\,\mathrm{Pa} \]

The adiabatic expansion cools the gas by 52 K because it does work at no heat cost; the free expansion leaves the temperature untouched because the gas does no work and exchanges no heat.

Review

Chapter Summary

✓ Avogadro & moles

\(N_{A} = 6.02\times10^{23}\,\mathrm{mol^{-1}}\), \(n = N/N_{A} = M_{\text{sam}}/M\), \(M = mN_{A}\).

✓ Ideal gas law

\(pV = nRT = NkT\), with \(R = 8.31\,\mathrm{J/mol\cdot K}\), \(k = 1.38\times10^{-23}\,\mathrm{J/K}\).

✓ Work by a gas

Isothermal \(W = nRT\ln(V_{f}/V_{i})\); const-p \(W = p\Delta V\); const-V \(W = 0\).

✓ Pressure & rms speed

\(p = nMv_{\text{rms}}^{2}/3V\), \(v_{\text{rms}} = \sqrt{3RT/M}\).

✓ Kinetic energy

\(K_{\text{avg}} = \tfrac{3}{2}kT\) — temperature is average translational KE.

✓ Mean free path

\(\lambda = 1/(\sqrt{2}\,\pi d^{2}\,N/V)\) — shrinks with crowding and molecular size.

✓ Speed distribution

Maxwell's \(P(v)\); \(v_{P} < v_{\text{avg}} < v_{\text{rms}}\).

✓ Molar specific heats

\(C_{V} = \tfrac{f}{2}R\), \(C_{p} = C_{V} + R\), \(\Delta E_{\text{int}} = nC_{V}\Delta T\).

✓ Adiabatic process

\(pV^{\gamma} = \text{const}\), \(TV^{\gamma-1} = \text{const}\), \(\gamma = C_{p}/C_{V}\).

Practice

Problems

Take \(R = 8.31\,\mathrm{J/mol\cdot K}\), \(N_{A} = 6.02\times10^{23}\,\mathrm{mol^{-1}}\), and \(1\,\mathrm{atm} = 1.01\times10^{5}\,\mathrm{Pa}\). Always convert temperatures to kelvins. For specific-heat problems, decide first whether the gas is monatomic (\(f=3\)) or diatomic (\(f=5\)), and remember \(\Delta E_{\text{int}} = nC_{V}\Delta T\) holds for every path.

Gold has a molar mass of 197 g/mol. In a 2.50 g sample, find (a) the number of moles and (b) the number of atoms.
A quantity of ideal gas at 10.0 °C and 100 kPa occupies 2.50 m³. (a) How many moles are present? (b) If the pressure is raised to 300 kPa and the temperature to 30.0 °C, what volume does it occupy?
An automobile tire has a volume of \(1.64\times10^{-2}\,\mathrm{m^{3}}\) and holds air at a gauge pressure of 165 kPa at 0.00 °C. Find the gauge pressure when the temperature rises to 27.0 °C and the volume grows to \(1.67\times10^{-2}\,\mathrm{m^{3}}\). (Atmospheric pressure 101 kPa.)
Suppose 1.80 mol of an ideal gas is compressed isothermally at 30 °C from 3.00 m³ to 1.50 m³. (a) How much energy is transferred as heat, and (b) is it to or from the gas?
Compute the rms speed of a nitrogen molecule (N₂, \(M = 28.0\,\mathrm{g/mol}\)) at 20.0 °C. At what temperatures is the rms speed (b) half and (c) twice that value?
Determine the average translational kinetic energy of an ideal-gas molecule at (a) 0.00 °C and (b) 100 °C, and (c) the kinetic energy per mole at 0.00 °C.
The mean free path of nitrogen at 0.0 °C and 1.0 atm is \(0.80\times10^{-5}\,\mathrm{cm}\), with \(2.7\times10^{19}\) molecules/cm³. What is the molecular diameter?
The speeds of ten molecules are 2.0, 3.0, 4.0, …, 11 km/s. Find their (a) average speed and (b) rms speed.
What is the internal energy of 1.0 mol of an ideal monatomic gas at 273 K?
The temperature of 3.00 mol of an ideal diatomic gas (rotation, no oscillation) rises 40.0 C° at constant pressure. Find (a) the heat added, (b) the change in internal energy, (c) the work done, and (d) the increase in rotational kinetic energy.
Under constant pressure, the temperature of 2.00 mol of an ideal monatomic gas is raised 15.0 K. Find (a) the work \(W\), (b) the heat \(Q\), and (c) the change \(\Delta E_{\text{int}}\).
We give 70 J as heat to a diatomic gas (rotation, no oscillation), which expands at constant pressure. By how much does its internal energy increase?
A certain gas occupies 4.3 L at 1.2 atm and 310 K, then is compressed adiabatically to 0.76 L (\(\gamma = 1.4\)). Find (a) the final pressure and (b) the final temperature.
An ideal monatomic gas at 330 K and 6.00 atm expands from 500 cm³ to 1500 cm³. If isothermal, find (a) the final pressure and (b) the work done; if adiabatic, find (c) the final pressure and (d) the work done.

Tip: three habits carry most of this chapter. For state problems, write \(pV = nRT\) for each state and take ratios so unknowns cancel — but keep temperature in kelvins. For energy problems, anchor on \(\Delta E_{\text{int}} = nC_{V}\Delta T\) (every path) and the first law \(\Delta E_{\text{int}} = Q - W\). For molecular questions, link macro to micro through \(K_{\text{avg}} = \tfrac{3}{2}kT\) and \(v_{\text{rms}} = \sqrt{3RT/M}\), recalling that lighter molecules move faster at the same temperature.