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Gas Laws and KMT
Kinetic Molecular Theory -- Postulates
The kinetic theory explains gas behavior through a microscopic model based on Elastic Collision and random molecular motion.
Large Number: A finite volume of gas consists of a very large number of molecules.
Negligible Size: Molecular dimensions are much smaller than the average separation between molecules.
Random Motion: Gas molecules move randomly and change direction after every collision.
Elastic Collisions: Collisions between molecules and with container walls are perfectly elastic -- kinetic energy is conserved.
No Intermolecular Forces: Molecules exert no force on each other except during a collision.
Pressure of a Gas from KMT
Pressure is the Momentum transferred to container walls per second per unit area due to continuous molecular collisions.
$$P = \frac{1}{3}\rho \langle v^2 \rangle$$
Gas pressure equals one-third the product of density and mean square velocity.
$P$=Pressure of the gas(Pa)
$\rho$=Gas density ($mN/V$)(kg/m³)
$\langle v^2 \rangle$=Mean square velocity of molecules(m²/s²)
In terms of $N$ and $V$
→$P = \frac{mN}{3V}\langle v^2 \rangle = \frac{2}{3}\frac{N}{V}\langle \frac{1}{2}mv^2 \rangle$
Momentum Change: Each elastic collision with a wall reverses the velocity component, giving $\Delta p = 2mv_x$ per collision.
Collision Rate: A molecule bouncing between opposite faces separated by $l$ hits a given face $v_x / 2l$ times per second.
Force per Molecule: Rate of momentum transfer gives $F = mv_x^2 / l$ for one molecule on one face.
The 1/3 Factor: Since motion is equally likely in x, y, z directions, each component contributes $\frac{1}{3}$ of $\langle v^2 \rangle$.
Key Result: $P \propto \langle KE \rangle$ -- pressure is directly proportional to the average translational kinetic energy.
Temperature as Average Kinetic Energy
Absolute Temperature is directly proportional to the average translational Kinetic Energy of gas molecules -- temperature is the macroscopic face of microscopic motion.
$$\langle KE \rangle = \frac{3}{2}kT$$
Average translational kinetic energy per molecule equals (3/2)kT.
$\langle KE \rangle$=Average translational kinetic energy per molecule(J)
$k$=Boltzmann constant(J/K)
$T$=Absolute temperature(K)
$T = 0$ K
→All translational kinetic energy is zero -- molecules cease translational motion.
Doubling $T$
→$\langle KE \rangle$ doubles; $v_{rms}$ increases by $\sqrt{2}$.
Derivation: Equating $PV = NkT$ with $PV = \frac{2}{3}N\langle \frac{1}{2}mv^2 \rangle$ gives $T = \frac{2}{3k}\langle \frac{1}{2}mv^2 \rangle$.
Mass Independence: At the same $T$, all gas molecules (light or heavy) have the same average KE -- heavier molecules simply move slower.
RMS Speed: $v_{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$ where $M$ is molar mass.
Proportionality Shortcut: $v_{rms} \propto \sqrt{T/M}$ -- useful for comparing speeds of different gases or temperatures.
The Ideal Gas Law
The Ideal Gas Law unifies Pressure, Volume, Absolute Temperature, and Moles into a single equation of state.
$$PV = nRT$$
For n moles of an ideal gas, the product PV is proportional to absolute temperature.
$P$=Pressure(Pa or atm)
$V$=Volume(m³ or L)
$n$=Number of moles(mol)
$R$=Universal gas constant(8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))
$T$=Absolute temperature(K)
Per-molecule form
→$PV = NkT$ where $N$ is total number of molecules and $k = R/N_A$
STP
→$P = 1$ atm, $T = 273$ K → 1 mol occupies 22.4 L
Density Form: $\rho = PM/(RT)$ -- derived by substituting $n = m/M$ and $\rho = m/V$.
Dimensional Check: $[Pa][m^3] = [mol][J/(mol \cdot K)][K] = [J]$ ✓
Derivation of Gas Laws from KMT
Boyle's Law: At constant temperature, Pressure and Volume are Inversely Proportional.
$$P_1V_1 = P_2V_2$$
From KMT: $PV = \frac{2}{3}N\langle \frac{1}{2}mv^2 \rangle$. At constant T, the right side is constant, so $PV = \text{const}$.
$P_1, V_1$=Initial pressure and volume(various)
$P_2, V_2$=Final pressure and volume(same as initial)
Volume halved
→Pressure doubles -- molecules hit walls twice as often in half the space.
Microscopic Reason: Smaller volume means shorter distance between walls, increasing collision frequency and hence pressure.
Process Name: An Isothermal process -- constant temperature.
Graph: $P$ vs $V$ is a rectangular hyperbola; $P$ vs $1/V$ is a straight line through the origin.
Charles's Law: At constant pressure, Volume is Directly Proportional to Absolute Temperature.
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
From KMT: $V = \frac{2N}{3P}\langle \frac{1}{2}mv^2 \rangle$. At constant P, $V \propto \langle KE \rangle \propto T$.
$V_1, T_1$=Initial volume and temperature(L, K)
$V_2, T_2$=Final volume and temperature(L, K)
Temperature doubled
→Volume doubles -- faster molecules push the piston out to maintain constant pressure.
Microscopic Reason: Higher $T$ means higher $\langle KE \rangle$; molecules hit walls harder, so volume must expand to keep $P$ constant.
Process Name: An Isobaric process -- constant pressure.
Internal Energy
Internal Energy ($U$) is the sum of all molecular kinetic and potential energies. For an ideal gas, it is entirely translational kinetic energy.
$$U = \frac{3}{2}nRT$$
For a monatomic ideal gas, internal energy equals (3/2)nRT since there is no potential energy between molecules.
$U$=Internal energy of the gas(J)
$n$=Number of moles(mol)
$R$=Universal gas constant (8.314 J/(mol·K))(J/(mol·K))
$T$=Absolute temperature(K)
Ideal gas
→$U$ depends ONLY on temperature, not on pressure or volume.
State Function: Internal energy depends only on the state (initial and final $T$), not on the path taken between states.
Energy Addition: Internal energy can increase via heat transfer OR mechanical work (e.g., friction, compression).
Diatomic Molecules: Have additional rotational and vibrational energy -- $U = \frac{5}{2}nRT$ at moderate temperatures.
Work Done by a Gas
When a gas expands against a piston, it does Work on the surroundings. Work done by the system is positive; work done on the system is negative.
$$W = P\Delta V$$
At constant pressure, work equals pressure times the change in volume.
$W$=Work done by the gas(J)
$P$=Constant pressure(Pa)
$\Delta V$=Change in volume ($V_f - V_i$)(m³)
$\Delta V = 0$ (constant volume)
→$W = 0$ -- no work done. All heat goes to internal energy.
Expansion ($\Delta V > 0$)
→$W > 0$ -- gas does positive work on surroundings.
Compression ($\Delta V < 0$)
→$W < 0$ -- surroundings do work on the gas.
PV Diagram: Work equals the area under the curve on a P-V graph.
Sign Convention: Heat IN is positive ($+Q$), work OUT by the system is positive ($+W$).
First Law: $Q = \Delta U + W$ -- heat added equals change in internal energy plus work done by the gas.
Molar Specific Heats of a Gas
Gases have two Molar Specific Heat values: $C_v$ (at constant Volume) and $C_p$ (at constant Pressure), because heating at constant pressure also requires work for expansion.
$$C_p - C_v = R$$
The difference between molar specific heats at constant pressure and constant volume equals the universal gas constant.
$C_p$=Molar specific heat at constant pressure(J/(mol·K))
$C_v$=Molar specific heat at constant volume(J/(mol·K))
$R$=Universal gas constant (8.314 J/(mol·K))(J/(mol·K))
Monatomic ideal gas
→$C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$, $\gamma = C_p/C_v = 5/3$
Diatomic ideal gas (moderate T)
→$C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, $\gamma = 7/5 = 1.4$
Why $C_p > C_v$: At constant pressure, some heat goes into expansion work ($P\Delta V = R\Delta T$ for 1 mol), so more heat is needed for the same $\Delta T$.
At Constant Volume: $\Delta U = C_v \Delta T$ -- all heat goes to internal energy, no work done.
Derivation: From first law, $C_p \Delta T = C_v \Delta T + P\Delta V = C_v \Delta T + R\Delta T$, giving $C_p = C_v + R$.
Application Strategy
Worked Examples
Problem Statement
What is the average translational kinetic energy of molecules in a gas at 27°C?
KMTTemperature-KE relationUnit Conversion
Strategic Toolkit
$$\langle KE \rangle = \frac{3}{2}kT$$
Step 1
Convert Temperature
Convert Celsius to Kelvin.
$T = 27 + 273 = 300$ K
Step 2
Apply the Formula
Substitute $k = 1.38 \times 10^{-23}$ J/K and $T = 300$ K.
$\langle KE \rangle = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300$
Step 3
Calculate
Evaluate the product.
$\langle KE \rangle = 6.21 \times 10^{-21}$ J
Problem Statement
Find the RMS speed of an oxygen molecule at STP (0°C, 1 atm).
RMS SpeedMolecular MassAvogadro's Number
Strategic Toolkit
$$v_{rms} = \sqrt{\frac{3kT}{m}}$$
Step 1
Find Mass of One O₂ Molecule
Divide molar mass by Avogadro's number.
$m = \frac{32 \text{ g}}{6.022 \times 10^{23}} = \frac{32}{6.022 \times 10^{26}}$ kg
Step 2
Set Up the Formula
Use $T = 273$ K and $k = 1.38 \times 10^{-23}$ J/K.
$\langle v^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 273 \times 6.022 \times 10^{26}}{32}$
Step 3
Calculate
Evaluate to find the mean square speed, then take the square root.
$\langle v^2 \rangle = 212693$ m²/s², so $v_{rms} = \sqrt{212693} \approx 461$ m/s
Problem Statement
Calculate the volume of 0.50 moles of an ideal gas at 2.0 atm and 27°C.
Ideal Gas LawUnit Conversion
Strategic Toolkit
$$PV = nRT$$
Step 1
Temperature Conversion
Convert Celsius to Kelvin.
$T = 27 + 273 = 300$ K
Step 2
Isolate Variable
Rearrange to solve for Volume.
$V = \frac{nRT}{P}$
Step 3
Calculation
Substitute $n = 0.50$, $R = 0.0821$ L·atm/(mol·K), $T = 300$ K, $P = 2.0$ atm.
$V = \frac{0.50 \times 0.0821 \times 300}{2.0} \approx 6.16$ L
Problem Statement
A gas at 1.0 atm and 5.0 L is compressed to 2.0 L while heated from 200 K to 400 K. Find the final pressure.
Combined Gas Law
Strategic Toolkit
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
Step 1
Identify States
$P_1 = 1.0$ atm, $V_1 = 5.0$ L, $T_1 = 200$ K, $V_2 = 2.0$ L, $T_2 = 400$ K.
$P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}$
Step 2
Solve
Substitute and calculate.
$P_2 = \frac{1.0 \times 5.0 \times 400}{2.0 \times 200} = \frac{2000}{400} = 5.0$ atm
Problem Statement
Show that the difference between molar specific heats at constant pressure and constant volume equals R for an ideal gas.
First Law of ThermodynamicsMolar Specific Heats
Strategic Toolkit
$$Q = \Delta U + W$$
Step 1
Constant Volume Heating
At constant volume, $W = 0$, so $Q_v = \Delta U = C_v \Delta T$.
$\Delta U = C_v \Delta T$
Step 2
Constant Pressure Heating
At constant pressure, the gas also does expansion work $W = P\Delta V$.
$C_p \Delta T = C_v \Delta T + P\Delta V$
Step 3
Use Ideal Gas Law
For 1 mole at constant $P$: $P\Delta V = R\Delta T$.
$C_p \Delta T = C_v \Delta T + R\Delta T$
Step 4
Simplify
Divide through by $\Delta T$.
$C_p - C_v = R$
Memorize formulas
KMT Pressure
Pressure from kinetic theory -- one-third density times mean square speed.
Formula
$P = \frac{1}{3}\rho \langle v^2 \rangle$
Average KE per Molecule
Temperature is proportional to average translational kinetic energy.
Formula
$\langle KE \rangle = \frac{3}{2}kT$
Ideal Gas Law
The master equation relating P, V, n, and T.
Formula
PV = nRT
Boyle's Law
Pressure and Volume inversely related at constant T.
Formula
$P_1V_1 = P_2V_2$
Charles's Law
Volume and Temperature directly related at constant P.
Formula
$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
RMS Speed
Molecular speed from temperature and molar mass.
Formula
$v_{rms} = \sqrt{\frac{3RT}{M}}$
Cp − Cv Relation
Molar specific heat difference equals universal gas constant.
Formula
$C_p - C_v = R$
Work by Gas
Work done by gas at constant pressure.
Formula
$W = P\Delta V$
Review
1
The average translational KE of gas molecules at $127°C$ is $E_1$ and at $527°C$ is $E_2$. The ratio $E_2/E_1$ is:
Ans
$2$
Review
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Convert both temperatures to Kelvin
Average translational kinetic energy is proportional to absolute temperature, so the Celsius values must be shifted to Kelvin before comparing.
$$T_1 = 127 + 273 = 400\text{ K},\quad T_2 = 527 + 273 = 800\text{ K}$$
Use the direct proportionality
Because $\langle KE \rangle \propto T$, the energy ratio equals the temperature ratio.
$$\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{800}{400} = 2$$
2
In the KMT pressure derivation, the factor $\frac{1}{3}$ arises because:
Ans
Random molecular motion is equally distributed among three perpendicular directions
Review
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Use isotropic molecular motion
In an ideal gas, motion is equally likely in the three perpendicular directions, so the mean square speed is shared equally among the components.
$$\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle = 3\langle v_x^2 \rangle$$
3
Two moles of an ideal gas at constant pressure absorb $500$ J of heat, raising the temperature by $\Delta T$. The work done by the gas during this process is:
Ans
$\frac{500R}{C_p}$ J
Review
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Relate heat input to temperature rise
For a constant-pressure process, the absorbed heat is $Q = nC_p\Delta T$, so the temperature rise is fixed by the heat capacity.
$$500 = 2C_p\Delta T \;\Rightarrow\; \Delta T = \frac{500}{2C_p}$$
Compute the expansion work
At constant pressure, the work is $W = nR\Delta T$, so substitute the temperature rise found above.
$$W = 2R\left(\frac{500}{2C_p}\right) = \frac{500R}{C_p}$$
4
The RMS speed of gas molecules at $T$ K is $v$. If the temperature is increased to $4T$ and the gas dissociates so each molecule splits into two equal fragments, the new RMS speed is:
Ans
$2\sqrt{2}\, v$
Review
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Apply the RMS speed formula to the new gas fragments
The RMS speed depends on temperature and particle mass, so both the higher temperature and the reduced fragment mass must be included.
$$v' = \sqrt{\frac{3k(4T)}{m/2}} = \sqrt{8\cdot\frac{3kT}{m}}$$
Compare with the original speed
Since $v = \sqrt{3kT/m}$, the new speed is a factor of $\sqrt{8}$ times the original.
$$v' = \sqrt{8}\,v = 2\sqrt{2}\,v$$
5
If the RMS speed of a gas molecule is doubled, by what factor must the absolute temperature have changed?
Ans
Factor of $4$
Review
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Square the speed change
RMS speed is proportional to the square root of absolute temperature, so a factor change in speed must be squared to get the temperature change.
$$v_{rms} \propto \sqrt{T} \;\Rightarrow\; T \propto v_{rms}^2$$
Double the speed and evaluate
If the speed doubles, the temperature must change by the square of 2.
$$\left(2\right)^2 = 4$$
6
A gas expands from $V$ to $2V$ at constant pressure $P$, then from $2V$ to $4V$ at constant pressure $P/2$. The total work done by the gas is:
Ans
$2PV$
Review
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Compute the work for each isobaric leg
For a constant-pressure step, work is pressure times volume change. Evaluate each segment separately and then add them.
$$W_1 = P(2V-V)=PV,\quad W_2 = \frac{P}{2}(4V-2V)=PV$$
Add the two contributions
The total work is the sum of both stages of expansion.
$$W = W_1 + W_2 = PV + PV = 2PV$$
7
For a diatomic ideal gas at moderate temperature, the ratio $C_p/C_v$ equals:
Ans
$7/5$
Review
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Use the active degrees of freedom
A diatomic gas at moderate temperature has 5 active degrees of freedom, giving the standard molar heat capacity at constant volume.
$$C_v = \frac{5}{2}R$$
Add one gas constant for constant pressure
For any ideal gas, $C_p = C_v + R$, so the ratio follows immediately.
$$C_p = \frac{7}{2}R,\quad \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5}$$
8
An ideal gas at $300$ K occupies $10$ L at $1$ atm. It is compressed isothermally to $5$ L. The new pressure is:
Ans
$2$ atm
Review
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Apply Boyle's law
Because the compression is isothermal, pressure and volume remain inversely proportional, so the product $PV$ stays constant.
$$P_1V_1 = P_2V_2$$
Solve for the final pressure
Substitute the given values and isolate $P_2$.
$$P_2 = \frac{P_1V_1}{V_2} = \frac{(1\,\text{atm})(10\,\text{L})}{5\,\text{L}} = 2\,\text{atm}$$
9
The internal energy of an ideal gas depends on:
Ans
Temperature only
Review
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Identify what contributes to internal energy
An ideal gas has negligible intermolecular potential energy, so its internal energy comes only from molecular kinetic energy.
$$U = \frac{f}{2}nRT$$
Read off the variable dependence
Since $n$ and $f$ are fixed for the gas, the only changing variable is temperature.
$$U \text{ depends on temperature only}$$
10
One mole of an ideal gas at constant volume receives $Q$ joules of heat. The change in internal energy is:
Ans
$Q$
Review
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Use the first law of thermodynamics
The heat supplied is split between internal energy change and work done by the gas.
$$Q = \Delta U + W$$
Set the work to zero at fixed volume
Since the volume does not change, the gas does no expansion work, so all the heat becomes internal energy.
$$W = P\Delta V = 0 \;\Rightarrow\; \Delta U = Q$$
11
Two gases have equal RMS speeds. Gas A has molar mass $M$ and temperature $T$. Gas B has molar mass $4M$. The temperature of gas B must be:
Ans
$4T$
Review
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Set the RMS speed expressions equal
Equal speeds mean the temperature-to-molar-mass ratios must match for both gases.
$$\sqrt{\frac{3RT}{M}} = \sqrt{\frac{3RT_B}{4M}}$$
Solve for the unknown temperature
Square both sides and isolate $T_B$.
$$\frac{T}{M} = \frac{T_B}{4M} \;\Rightarrow\; T_B = 4T$$
Question 1 of 11
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1
The average translational KE of gas molecules at $127°C$ is $E_1$ and at $527°C$ is $E_2$. The ratio $E_2/E_1$ is: