Gas Laws and Kinetic Molecular Theory
Kinetic Molecular Theory — Postulates
The Kinetic Molecular Theory explains the macroscopic behaviour of gases through the microscopic motion of their molecules. It was first proposed by Bernoulli (1738) and later developed by Clausius (1857), Maxwell, and Boltzmann. The theory rests on eight fundamental postulates that describe an ideal gas.
Composition: Every gas consists of a large number of very small particles called molecules. Monoatomic gases such as He, Ne, and Ar have single-atom molecules.
Random Motion: Gas molecules move haphazardly, constantly colliding among themselves and with the walls of the container, changing direction after each collision.
Elastic Collisions: All collisions — between molecules and with the walls — are perfectly elastic. Kinetic energy is conserved during every collision.
Large Separation: Molecules of a gas are widely separated from one another, with sufficient empty spaces between them. Most of the volume occupied by a gas is actually empty space.
No Intermolecular Forces: The molecules of a gas exert no forces of attraction or repulsion on each other.
Negligible Molecular Volume: The actual volume of gas molecules is negligible compared to the total volume of the gas.
Gravity Negligible: The effect of gravity on molecular motion is negligible compared to the effect of continued collisions between molecules.
Kinetic Energy and Temperature: The average kinetic energy of gas molecules varies directly as the absolute temperature of the gas.
Eight Postulates at a Glance
1
Large number of very small molecules
2
Haphazard, random motion of molecules
3
Perfectly elastic collisions
4
Widely separated molecules with empty spaces
5
No intermolecular forces of attraction
6
Negligible molecular volume vs container volume
7
Gravity effect negligible compared to collisions
8
Average KE ∝ absolute temperature
Using these postulates, Clausius derived the Kinetic Equation for an ideal gas. This equation connects the macroscopic properties of pressure and volume to the microscopic properties of molecular mass and speed.
$$PV = \frac{1}{3} mN\overline{c^2}$$
Relates the pressure and volume of a gas to the mass, number, and mean square velocity of its molecules
$P$=Pressure exerted by the gas(atm or Pa)
$V$=Volume of the container(dm³ or m³)
$m$=Mass of one molecule of the gas(kg)
$N$=Total number of molecules in the vessel(dimensionless)
$\overline{c^2}$=Mean square velocity of the molecules(m²/s²)
Mean Square Velocity: All molecules of a gas do not have the same velocity. The quantity $\overline{c^2}$ is the average of the squares of all molecular velocities. Its square root gives the RMS Velocity.
RMS Formula: The root mean square velocity can also be expressed in terms of molar mass and temperature, showing that lighter molecules move faster at the same temperature.
Pressure Origin: Pressure arises from the force of molecular collisions with the container walls divided by the wall area — it is a macroscopic result of microscopic impacts.
The RMS Velocity is a useful measure of molecular speed that follows directly from the Kinetic Equation. It quantitatively links molecular speed to temperature and molar mass.
$$C_{rms} = \sqrt{\frac{3RT}{M}}$$
Gives the root mean square speed of gas molecules in terms of temperature and molar mass
$C_{rms}$=Root mean square velocity(m/s)
$R$=Universal gas constant(8.314 J mol⁻¹ K⁻¹)
$T$=Absolute temperature(K)
$M$=Molar mass of the gas(kg/mol)
$T$ increases
→$C_{rms}$ increases — molecules move faster at higher temperatures
$M$ increases
→$C_{rms}$ decreases — heavier molecules move slower at the same temperature
Temperature Dependence: At higher temperatures, molecules possess greater kinetic energy and therefore move at higher RMS velocities.
Molar Mass Dependence: For two gases at the same temperature, the lighter gas has the higher RMS velocity.
Gas Particle Motion and Pressure
Gas molecules are in a state of continuous, random motion. They travel in straight lines between collisions and change direction abruptly upon each collision. The Pressure a gas exerts on its container is entirely due to these collisions — each impact transfers momentum to the wall, and the summed effect of billions of collisions per second produces the macroscopic pressure we measure.
$$E_k = \frac{3R}{2N_A}T$$
Defines the average translational kinetic energy per molecule of a gas in terms of absolute temperature
$E_k$=Average translational kinetic energy per molecule(J)
$R$=Universal gas constant (8.314 J mol⁻¹ K⁻¹)(J mol⁻¹ K⁻¹)
$N_A$=Avogadro's number (6.022 × 10²³)(mol⁻¹)
$T$=Absolute temperature(K)
$T = 0$ K
→$E_k = 0$ — molecular motion would cease at absolute zero (unattainable in practice)
Elastic Collisions: Both molecule-molecule and molecule-wall collisions conserve total kinetic energy. No energy is lost as heat or deformation during collisions of ideal gas molecules.
Maxwell's Distribution: Molecular velocities are not identical. Molecules group into velocity ranges, with most molecules near the most probable speed. The distribution broadens and shifts to higher speeds as temperature increases.
KE and Temperature Link: From the kinetic equation, comparing $PV = RT$ for one mole with $PV = \frac{2}{3}N_A E_k$ gives $E_k = \frac{3R}{2N_A}T$. This confirms that temperature is directly proportional to the average translational kinetic energy of molecules.
Boyle's Law
Boyle's Law describes the inverse relationship between the pressure and volume of a fixed amount of gas at constant temperature. When you compress a gas (increase pressure), its volume decreases proportionally, and vice versa.
$$P_1 V_1 = P_2 V_2$$
At constant temperature, the product of pressure and volume for a given mass of gas is always constant
$P_1, V_1$=Initial pressure and volume(atm, dm³)
$P_2, V_2$=Final pressure and volume(atm, dm³)
Temperature doubles
→PV remains constant only if T is held fixed — changing T changes the constant k
Inverse Proportionality: $V \propto \frac{1}{P}$ when temperature and number of moles are constant. Doubling the pressure halves the volume.
Constant k: The product $PV = k$ is constant for the same quantity of gas at the same temperature. Different amounts of gas or different temperatures give different k values.
Experimental Basis: When a gas at 2 atm occupying 1 dm³ is compressed to 4 atm, its volume becomes $\frac{1}{2}$ dm³, and at 6 atm it becomes $\frac{1}{3}$ dm³. In each case $PV = 2$ dm³ atm.
Boyle's law can be represented graphically in three ways, each confirming the inverse proportionality between pressure and volume.
V vs P (Isotherm): A plot of volume against pressure gives a hyperbolic curve called an isotherm ('iso' = same, 'therm' = heat). At higher temperatures, the isotherm shifts away from both axes because the gas occupies a larger volume at each pressure.
P vs 1/V: A plot of pressure against $\frac{1}{V}$ gives a straight line passing through the origin. At higher temperatures, this line has a steeper slope (closer to the pressure axis).
PV vs P: A plot of the product $PV$ against pressure gives a horizontal straight line parallel to the pressure axis, confirming that $PV$ is constant at a given temperature.
Boyle's Law can be derived from the Kinetic Equation. Starting from $PV = \frac{2}{3} \times \frac{1}{2}mN\overline{c^2}$ and using the fact that average kinetic energy is proportional to temperature ($\frac{1}{2}mN\overline{c^2} = kT$), we get $PV = \frac{2}{3}kT$. At constant temperature, the right side is a constant, so $PV = k'$ — which is Boyle's law.
Molecular Reasoning: When volume decreases at constant temperature, molecules travel shorter distances between wall collisions and strike the walls more frequently per unit area — hence pressure increases.
Temperature Link: Since $T$ is constant, the average speed of molecules does not change. The pressure increase is purely due to more frequent collisions in a smaller volume.
Charles's Law
Charles's Law describes the direct relationship between the volume and absolute temperature of a fixed amount of gas at constant pressure. As a gas is heated, its volume increases proportionally, and as it is cooled, its volume decreases.
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
At constant pressure, the ratio of volume to absolute temperature for a given mass of gas is always constant
$V_1, T_1$=Initial volume and temperature(dm³, K)
$V_2, T_2$=Final volume and temperature(dm³, K)
$T = 0$ K
→Volume would theoretically become zero (absolute zero)
Direct Proportionality: $V \propto T$ when pressure and number of moles are constant. Doubling the absolute temperature doubles the volume.
Constant k: The ratio $\frac{V}{T} = k$ is constant for the same amount of gas at the same pressure.
Volume-Temperature Data: For a gas with volume 546 cm³ at 0 °C (273 K), the ratio $\frac{V}{T} = 2$ cm³/K remains constant whether the temperature is 100 °C (373 K, $V$ = 746 cm³) or −100 °C (173 K, $V$ = 346 cm³).
Charles's Law follows from the Kinetic Equation. From $PV = \frac{2}{3}kT$, at constant pressure we get $V = \frac{2k}{3P}T$. Since $\frac{2k}{3P}$ is a constant when $P$ is fixed, we have $V = k''T$ or $\frac{V}{T} = k''$ — which is Charles's law.
Molecular Reasoning: At constant pressure, when temperature increases, molecules move faster and push the piston outward with greater force, expanding the volume until the internal pressure matches the external pressure again.
Isothermal vs Isobaric: Boyle's law describes isothermal changes (constant T), while Charles's law describes isobaric changes (constant P).
Absolute Zero
Absolute Zero is the lowest possible temperature, theoretically the point where all molecular motion ceases. It is derived by extrapolating the volume-temperature relationship of a gas to the point where volume reaches zero.
$$V_t = V_0 \left(1 + \frac{t}{273}\right)$$
Gives the volume of a gas at any Celsius temperature, based on its volume at 0 °C
$V_t$=Volume of gas at temperature $t$ °C(cm³ or dm³)
$V_0$=Volume of gas at 0 °C(cm³ or dm³)
$t$=Temperature on Celsius scale(°C)
$t = -273$ °C
→$V_t = 0$ — the theoretical point of zero volume, defining absolute zero
1/273 Rule: At constant pressure, the volume of a given mass of gas changes by $\frac{1}{273}$ of its volume at 0 °C for every 1 °C change in temperature.
Graphical Derivation: A plot of volume (y-axis) against Celsius temperature (x-axis) gives a straight line. Extrapolating this line backward, it intersects the temperature axis at −273.16 °C (0 K), where volume would be zero.
Unattainable: Absolute zero (0 K or −273.16 °C) cannot actually be reached because all real gases liquefy before this temperature. It represents a theoretical limit.
Celsius Scale Failure: Charles's law is not obeyed when using Celsius temperatures. For example, $\frac{566}{10} \neq \frac{746}{100}$ in Celsius, but in Kelvin: $\frac{566}{283} = \frac{746}{373} = 2$.
Temperature Scales and STP
Three temperature scales are commonly used in chemistry. The Kelvin Scale (absolute scale) is the standard for gas law calculations because it starts from Absolute Zero and avoids negative values.
Centigrade (Celsius) Scale: Ice melts at 0 °C and water boils at 100 °C at 1 atm. The range is divided into 100 equal parts.
Fahrenheit Scale: Ice melts at 32 °F and water boils at 212 °F at 1 atm. The range is divided into 180 equal parts.
Kelvin (Absolute) Scale: The lowest point is 0 K (absolute zero). Ice melts at 273 K and water boils at 373 K (more precisely 373.16 K) at 1 atm.
Temperature Scale Conversion Formulas
1
$K = °C + 273.16$
2
$°C = \frac{5}{9}(°F - 32)$
3
$°F = \frac{9}{5}(°C) + 32$
Standard Temperature and Pressure (STP) is the reference condition used for comparing gas volumes. At STP, one mole of any ideal gas occupies exactly 22.4 dm³.
Standard Temperature: 0 °C (273 K)
Standard Pressure: 1 atmosphere (101.325 kPa or 760 mmHg)
Molar Volume: At STP, 1 mole of any ideal gas = 22.4 dm³. This is a fixed value that applies universally to all ideal gases regardless of their identity.
Practical Use: STP allows direct comparison of gas volumes and simplifies stoichiometric calculations involving gases.
STP Conditions Summary
•
Temperature: 0 °C (273 K)
•
Pressure: 1 atm (101.325 kPa, 760 mmHg)
•
Molar volume of ideal gas: 22.4 dm³/mol
Application Strategy
Worked Examples
Problem Statement
A gas having a volume of 10 dm³ is enclosed in a vessel at 0 °C and the pressure is 2.5 atm. The gas is allowed to expand until the new pressure is 2 atm. What will be the new volume if the temperature is maintained at 273 K?
Boyle's Law
Strategic Toolkit
$$P_1 V_1 = P_2 V_2$$
Step 1
Identify the given quantities and condition
Since temperature is constant (273 K throughout), Boyle's law applies. Initial pressure is 2.5 atm, initial volume is 10 dm³, and final pressure is 2 atm.
$P_1 = 2.5 \text{ atm},\; V_1 = 10 \text{ dm³},\; P_2 = 2 \text{ atm},\; V_2 = \text{?}$
Step 2
Apply Boyle's law
Rearrange $P_1 V_1 = P_2 V_2$ to solve for the unknown final volume.
$$V_2 = \frac{P_1 V_1}{P_2}$$
Step 3
Substitute and calculate
Substitute the known values to find the new volume. The units of atm cancel out.
$$V_2 = \frac{2.5 \text{ atm} \times 10 \text{ dm³}}{2 \text{ atm}} = 12.5 \text{ dm³}$$
Problem Statement
250 cm³ of hydrogen is cooled from 127 °C to −27 °C while maintaining the pressure constant. Calculate the new volume of the gas.
Charles's Law
Strategic Toolkit
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
Step 1
Convert temperatures to Kelvin
Charles's law requires absolute temperature. Convert both Celsius temperatures to Kelvin by adding 273.
$T_1 = 127 + 273 = 400 \text{ K},\quad T_2 = -27 + 273 = 246 \text{ K}$
Step 2
Apply Charles's law
Since pressure is constant, use the Charles's law equation to find the final volume.
$$V_2 = \frac{V_1 \times T_2}{T_1} = \frac{250 \text{ cm³} \times 246 \text{ K}}{400 \text{ K}}$$
Step 3
Calculate the final volume
The temperature decreased, so the volume should also decrease — from 250 cm³ down to 153.75 cm³.
$$V_2 = \frac{250 \times 246}{400} = \frac{61500}{400} = 153.75 \text{ cm³} \approx 154 \text{ cm³}$$
Problem Statement
A gas occupies 3.0 dm³ at 2 atm and 300 K. If the pressure is changed to 3 atm and the temperature to 450 K, what is the new volume?
Boyle's LawCharles's Law
Strategic Toolkit
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
Step 1
Identify all conditions
Both pressure and temperature change, so we need the combined gas equation. List all known and unknown quantities.
$P_1 = 2 \text{ atm},\; V_1 = 3.0 \text{ dm³},\; T_1 = 300 \text{ K},\; P_2 = 3 \text{ atm},\; T_2 = 450 \text{ K},\; V_2 = \text{?}$
Step 2
Rearrange the combined equation
Solve for $V_2$ by rearranging the combined gas equation.
$$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$$
Step 3
Substitute and compute
Substitute all values. Pressure increases (tending to decrease volume) but temperature also increases (tending to increase volume). The net effect gives a volume slightly larger than the original.
$$V_2 = \frac{2 \text{ atm} \times 3.0 \text{ dm³} \times 450 \text{ K}}{3 \text{ atm} \times 300 \text{ K}} = \frac{2700}{900} = 3.0 \text{ dm³}$$
Problem Statement
A gas has a volume of 546 cm³ at 0 °C. According to Charles's law, at what Celsius temperature would the volume of this gas theoretically become zero?
Charles's LawAbsolute Zero
Strategic Toolkit
$$V_t = V_0 \left(1 + \frac{t}{273}\right)$$
Step 1
Set volume to zero and solve
To find the temperature at which volume becomes zero, set $V_t = 0$ in the Charles's law volume-temperature equation.
$$0 = 546 \left(1 + \frac{t}{273}\right)$$
Step 2
Solve for t
Since $V_0 = 546$ cm³ is non-zero, the term in parentheses must be zero for the product to be zero.
$$1 + \frac{t}{273} = 0 \implies \frac{t}{273} = -1 \implies t = -273 \text{ °C}$$
Problem Statement
Calculate the RMS velocity of oxygen molecules ($M = 32$ g/mol) at 300 K. ($R = 8.314$ J mol⁻¹ K⁻¹)
RMS Velocity
Strategic Toolkit
$$C_{rms} = \sqrt{\frac{3RT}{M}}$$
Step 1
Convert molar mass to kg/mol
The SI unit for molar mass in the RMS formula is kg/mol, so convert 32 g/mol to 0.032 kg/mol.
$M = 32 \text{ g/mol} = 0.032 \text{ kg/mol}$
Step 2
Substitute into the RMS formula
Insert the values of R, T, and M into the RMS velocity equation and evaluate.
$$C_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233831.25} \approx 483.6 \text{ m/s}$$
Memorize formulas
Boyle's Law
Use when temperature and moles are constant. P and V are inversely proportional. Units: atm and dm³ (or any consistent pressure-volume pair).
Formula
$$P_1 V_1 = P_2 V_2$$
Charles's Law
Use when pressure and moles are constant. V and T (Kelvin) are directly proportional. Always convert °C to K first.
Formula
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
Kinetic Equation
Connects macroscopic PV to microscopic molecular properties. m = mass of one molecule, N = number of molecules, $\overline{c^2}$ = mean square velocity.
Formula
$$PV = \frac{1}{3} mN\overline{c^2}$$
RMS Velocity
R = 8.314 J mol⁻¹ K⁻¹, M in kg/mol, T in K. Lighter gases have higher RMS velocity at the same T.
Formula
$$C_{rms} = \sqrt{\frac{3RT}{M}}$$
Volume-Temperature Equation
Gives gas volume at any Celsius temperature. V₀ = volume at 0 °C. Setting Vₜ = 0 gives t = −273 °C (absolute zero).
Formula
$$V_t = V_0 \left(1 + \frac{t}{273}\right)$$
Average Translational KE per Molecule
R = 8.314 J mol⁻¹ K⁻¹, N_A = 6.022 × 10²³ mol⁻¹. Shows temperature is proportional to average KE of molecules.
Formula
$$E_k = \frac{3R}{2N_A}T$$
Celsius to Kelvin Conversion
Use for all gas law calculations. Kelvin scale starts from absolute zero (0 K = −273.16 °C).
Formula
$$K = °C + 273$$
Review
1
According to the kinetic molecular theory, collisions between gas particles are:
Ans
Perfectly elastic — total kinetic energy is conserved
Review
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Recall KMT postulates
KMT postulates state that gas particle collisions are perfectly elastic with no net loss of kinetic energy.
Elastic collisions conserve total kinetic energy
2
Which statement correctly describes the kinetic molecular theory's postulate about intermolecular forces between gas particles?
Ans
No significant intermolecular forces exist between gas particles except during brief collisions
Review
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Identify the relevant KMT postulate
KMT assumes ideal gas behavior where intermolecular forces are absent except during instantaneous collisions.
No significant intermolecular forces between particles
3
What is the origin of gas pressure according to the kinetic molecular theory?
Ans
Force per unit area produced by collisions of particles with the container walls
Review
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Link particle motion to pressure
Gas particles move randomly and collide with walls. Each collision imparts momentum, generating force.
$$P = \frac{F}{A}$$
Connect to kinetic equation
The kinetic equation $PV = \frac{1}{3}mN\overline{c^2}$ derives pressure from particle mass, number, and mean square velocity.
Pressure = force per unit area from collisions
4
The root mean square velocity of a gas molecule depends on which of the following?
Ans
Temperature and molar mass only
Review
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Write the RMS velocity formula
From kinetic molecular theory, the RMS velocity is derived as:
$$C_{rms} = \sqrt{\frac{3RT}{M}}$$
Identify the variables
$R$ is the universal gas constant, $T$ is absolute temperature, and $M$ is the molar mass.
$C_{rms}$ depends on $T$ and $M$ only
5
Boyle's law states that at constant temperature, the volume of a fixed mass of gas is:
Ans
Inversely proportional to pressure
Review
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State Boyle's law
For a fixed mass of gas at constant temperature:
$$P_1 V_1 = P_2 V_2$$
Express the proportionality
Since $PV = \text{constant}$, volume is inversely proportional to pressure.
$V \propto \frac{1}{P}$
6
A gas occupies 500 cm³ at a pressure of 760 torr. If the pressure is increased to 1140 torr at constant temperature, what is the new volume?
Ans
333 cm³
Review
Tap to reveal
Apply Boyle's law
At constant temperature, $P_1 V_1 = P_2 V_2$.
$$P_1 V_1 = P_2 V_2$$
Substitute values
$760 \times 500 = 1140 \times V_2$
$$V_2 = \frac{380000}{1140} = 333 \text{ cm}^3$$
7
On a pressure-volume graph for a gas at constant temperature, the plotted curve is:
Ans
A rectangular hyperbola where pressure is inversely proportional to volume
Review
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Write Boyle's law equation
At constant temperature, $PV = k$ (constant).
$$P = \frac{k}{V}$$
Identify the graph shape
The equation $P = k/V$ is the equation of a rectangular hyperbola — the defining curve of an isotherm.
Rectangular hyperbola (Boyle's law isotherm)
8
Charles's law describes the relationship between which two variables at constant pressure?
Ans
Volume and temperature
Review
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State Charles's law
At constant pressure for a fixed mass of gas:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
Identify the variables
Volume and absolute temperature are the two related variables.
$V \propto T$ (at constant $P$)
9
A gas sample has a volume of 300 cm³ at 27°C. If the pressure remains constant, what will be its volume at 127°C?
Ans
400 cm³
Review
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Convert temperatures to Kelvin
$T_1 = 27 + 273 = 300$ K and $T_2 = 127 + 273 = 400$ K.
$$T_1 = 300 \text{ K}, \quad T_2 = 400 \text{ K}$$
Apply Charles's law
$V_2 = V_1 \times \frac{T_2}{T_1} = 300 \times \frac{400}{300}$
$$V_2 = 400 \text{ cm}^3$$
10
A gas occupies 2.0 dm³ at STP (273 K, 1 atm). If the pressure is doubled and the temperature is increased to 546 K, what is the new volume?
Ans
2.0 dm³
Review
Tap to reveal
Apply Boyle's law for pressure change
Pressure doubles from 1 atm to 2 atm, so volume halves: $V' = 2.0 / 2 = 1.0$ dm³.
$$V' = \frac{V_1 \times P_1}{P_2} = 1.0 \text{ dm}^3$$
Apply Charles's law for temperature change
Temperature doubles from 273 K to 546 K, so volume doubles: $V_2 = 1.0 \times (546/273) = 2.0$ dm³.
$$V_2 = V' \times \frac{T_2}{T_1} = 2.0 \text{ dm}^3$$
11
Absolute zero is defined as the temperature at which:
Ans
Theoretically, all molecular motion ceases and volume reaches zero
Review
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Recall the definition of absolute zero
From the volume-temperature equation $V_t = V_0(1 + t/273)$, setting volume to zero gives $t = -273$°C.
$$0 \text{ K} = -273\degree C$$
Interpret physically
At absolute zero, gas particles would have zero kinetic energy and zero volume (theoretically for an ideal gas).
All molecular motion ceases
12
A gas has a volume of 546 cm³ at 0°C. According to Charles's law, at what temperature will the volume become 273 cm³ at the same pressure?
Ans
−136.5°C
Review
Tap to reveal
Convert initial temperature to Kelvin
$T_1 = 0 + 273 = 273$ K.
$$T_1 = 273 \text{ K}$$
Apply Charles's law
$\frac{546}{273} = \frac{273}{T_2}$, so $T_2 = \frac{273 \times 273}{546} = 136.5$ K.
$$T_2 = 136.5 \text{ K} = -136.5\degree C$$
13
Standard Temperature and Pressure (STP) is defined as:
Ans
0°C and 760 torr
Review
Tap to reveal
Recall STP conditions
Standard Temperature and Pressure is universally defined as 0°C (273 K) and 760 torr (1 atm).
$$\text{STP} = 0\degree C \text{ (273 K)}, \; 760 \text{ torr (1 atm)}$$
14
A gas is at 50°C. If it is heated to 150°C, what is the new temperature in Kelvin?
Ans
423 K
Review
Tap to reveal
Apply Celsius to Kelvin conversion
$K = \degree C + 273$
$$T = 150 + 273 = 423 \text{ K}$$
15
According to the kinetic molecular theory, the average translational kinetic energy per molecule of a gas depends on:
Ans
The absolute temperature only
Review
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Write the kinetic energy expression
The average translational kinetic energy per molecule is:
$$E_k = \frac{3R}{2N_A}T$$
Identify the variable
Since $R$, $N_A$, and the fraction $3/2$ are all constants, $E_k$ depends only on $T$.
$E_k \propto T$ (independent of molar mass)
Question 1 of 15
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1
According to the kinetic molecular theory, collisions between gas particles are: