Kinetic Theory of Matter and Pressure of Gases Practice Questions — JAMB UTME

Question 1Question →

A gas sample enclosed in a container has an initial root-mean-square (r.m.s.) speed of $400\text{ m s}^{-1}$ at a temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until its temperature reaches $927^\circ\text{C}$ , what is the new r.m.s. speed of the gas molecules?

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

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

1600\text{ m s}^{-1}

2400\text{ m s}^{-1}

Show answer & explanation

Answer: $800\text{ m s}^{-1}$

Answer

The new root-mean-square speed of the gas molecules is

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

.

According to the kinetic theory of gases, the root-mean-square speed is directly proportional to the square root of the absolute temperature (

v_{\text{rms}} = \sqrt{3RT/M}

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 1200\text{ K}

. The ratio of absolute temperatures is

1200 / 300 = 4

. Taking the square root gives a factor of

2

, so the new r.m.s. speed is

400\text{ m s}^{-1} \times 2 = 800\text{ m s}^{-1}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to Kelvin

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

and

T_2 = 927 + 273 = 1200\text{ K}

Kinetic theory equations require absolute temperature in Kelvin.

2

Apply the relationship between r.m.s. speed and absolute temperature

v_{\text{rms}} \propto \sqrt{T} \implies \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

The mean kinetic energy of gas molecules is directly proportional to absolute temperature.

3

Substitute the values and calculate the final speed

v_2

v_2 = 400 \times \sqrt{\frac{1200}{300}} = 400 \times \sqrt{4} = 400 \times 2 = 800\text{ m s}^{-1}

Evaluating the square root factor yields the updated r.m.s. speed.

Key Concept

Root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Estimated Time:1m 0s

Question 2Question →

According to the kinetic theory of matter, what causes the pressure exerted by a gas on the walls of its container?

The continuous elastic collisions of gas molecules with the walls of the container

The attractive forces pulling gas molecules towards the container walls

The heat capacity of individual gas molecules transferring thermal mass to the walls

The static expansion of fixed gas molecules taking up space

Show answer & explanation

Answer: The continuous elastic collisions of gas molecules with the walls of the container

Answer

The pressure exerted by a gas is due to the continuous elastic collisions of gas molecules with the container walls.

According to the kinetic theory of gases, gas particles are in rapid, random motion. When these particles collide elastically with the walls of the container, they undergo a change in momentum. The average rate of momentum change per unit area exerted by countless particle impacts manifests as macroscopic gas pressure.

Step-by-Step Solution

1

Identify the basic postulate of the kinetic theory of gases regarding particle motion.

Gas molecules are in rapid, constant, and random motion.

Kinetic theory models gas behavior based on particle movement.

2

Relate particle motion to force and pressure on the container boundary.

When particles hit the container wall, they undergo a change in momentum, imparting a force on the wall.

Force is defined as the rate of change of momentum (

F = \frac{\Delta p}{\Delta t}

).

3

Define pressure in terms of force per unit surface area.

The total force exerted per unit surface area by continuous collisions results in gas pressure (

P = \frac{F}{A}

).

Pressure is force distributed over a given surface area.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

Question 3Question →

A sample of gas enclosed in a rigid container exerts a pressure of $3.0 \times 10^5 \text{ N m}^{-2}$ with a density of $0.40 \text{ kg m}^{-3}$ . If the gas is heated at constant volume until its pressure increases to $1.2 \times 10^6 \text{ N m}^{-2}$ , what is the final root-mean-square (r.m.s.) speed of the gas molecules?

3000 \text{ m s}^{-1}

6000 \text{ m s}^{-1}

1500 \text{ m s}^{-1}

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

Show answer & explanation

Answer: $3000 \text{ m s}^{-1}$

Answer

The final root-mean-square speed of the gas molecules is

3000 \text{ m s}^{-1}

.

Using the kinetic theory equation

P = \frac{1}{3}\rho v_{\text{rms}}^2

, the initial speed is

v_1 = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.40}} = 1500 \text{ m s}^{-1}

. When heated at constant volume, density remains unchanged. The new pressure

1.2 \times 10^6 \text{ N m}^{-2}

is 4 times the initial pressure, so the new speed is

v_2 = \sqrt{4} \times v_1 = 2 \times 1500 \text{ m s}^{-1} = 3000 \text{ m s}^{-1}

.

Step-by-Step Solution

1

Express the relationship between pressure, density, and root-mean-square speed using kinetic theory.

P = \frac{1}{3}\rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

According to the kinetic theory of gases, the pressure exerted by a gas is related to its density and molecular r.m.s. speed.

2

Calculate the initial root-mean-square speed

v_1

using initial pressure

P_1 = 3.0 \times 10^5 \text{ N m}^{-2}

and density

\rho = 0.40 \text{ kg m}^{-3}

.

v_1 = \sqrt{\frac{3 \times (3.0 \times 10^5)}{0.40}} = \sqrt{\frac{9.0 \times 10^5}{0.40}} = \sqrt{2.25 \times 10^6} = 1500 \text{ m s}^{-1}

Establishes the baseline molecular speed prior to heating.

3

Determine the final root-mean-square speed

v_2

after the pressure increases to

P_2 = 1.2 \times 10^6 \text{ N m}^{-2}

at constant volume.

v_2 = \sqrt{\frac{3 \times (1.2 \times 10^6)}{0.40}} = \sqrt{\frac{3.6 \times 10^6}{0.40}} = \sqrt{9.0 \times 10^6} = 3000 \text{ m s}^{-1}

Since the volume is rigid, density

\rho

remains constant while pressure increases due to heating.

Key Concept

Root-Mean-Square Speed and Pressure Relation in Kinetic Theory

Question 4Question →

An ideal gas occupies a volume of $0.05\text{ m}^3$ inside a rigid container. If the gas exerts a pressure of $2.4 \times 10^5\text{ N m}^{-2}$ on the walls of the container, what is the total translational kinetic energy of the gas molecules in joules?

Show answer & explanation

Answer: 18000

Answer

The total translational kinetic energy of the gas molecules is

18000\text{ J}

.

According to the kinetic theory of gases, the pressure

P

of an ideal gas is related to its total translational kinetic energy

E_k

and volume

V

by

P = \frac{2}{3} \frac{E_k}{V}

. Rearranging for

E_k

gives

E_k = \frac{3}{2} P V

. Substituting

P = 2.4 \times 10^5\text{ N m}^{-2}

and

V = 0.05\text{ m}^3

yields

E_k = \frac{3}{2} \times 2.4 \times 10^5 \times 0.05 = 18000\text{ J}

.

Step-by-Step Solution

1

Identify the relationship between gas pressure, volume, and translational kinetic energy from kinetic theory.

P = \frac{2}{3} \left(\frac{E_k}{V}\right) \implies E_k = \frac{3}{2} P V

From kinetic theory, pressure is two-thirds of the total translational kinetic energy per unit volume.

2

Substitute the provided numerical values into the equation.

E_k = \frac{3}{2} \times (2.4 \times 10^5\text{ N m}^{-2}) \times (0.05\text{ m}^3)

The given values are pressure

P = 2.4 \times 10^5\text{ N m}^{-2}

and volume

V = 0.05\text{ m}^3

.

3

Evaluate the expression to determine the numerical result.

E_k = 1.5 \times 12000 = 18000\text{ J}

Multiplying the values gives the energy in Joules.

Key Concept

Relationship between pressure, volume, and total translational kinetic energy of gas molecules (

E_k = \frac{3}{2} P V

).

Question 5Question →

Match each kinetic theory concept on the left with its correct physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Temperature matches with the measure of average translational kinetic energy; Pressure matches with the average force per unit area exerted by colliding gas molecules on container walls; Root-mean-square speed matches with the square root of the mean of squared speeds.

Temperature measures average translational kinetic energy per particle. Pressure originates from force per unit area due to elastic wall collisions. Root-mean-square speed is the square root of the mean of squared molecular speeds.

Step-by-Step Solution

1

Identify the kinetic theory definition of Temperature

Temperature is directly proportional to the mean translational kinetic energy of the gas particles (

E_k \propto T

).

Absolute temperature reflects the average kinetic energy of molecular motion.

2

Identify the microscopic origin of Gas Pressure

Pressure is caused by molecular collisions with the container walls, transferring momentum and creating force per unit area.

Frequent elastic collisions of particles on container walls produce measurable pressure.

3

Identify the mathematical definition of Root-Mean-Square Speed

v_{rms} = \sqrt{\overline{v^2}}

, representing the square root of the average of squared molecular velocities.

This parameter represents the effective speed of gas particles relevant to thermal kinetic energy.

Key Concept

Kinetic Theory Interpretation of Gas Properties

Question 6Question →

Match each kinetic theory parameter of an ideal gas on the left with its corresponding microscopic physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Temperature corresponds to the measure of average translational kinetic energy; Gas pressure corresponds to the rate of momentum transfer per unit area from wall collisions; Root-mean-square speed corresponds to the square root of the mean of squared speeds.

Temperature is directly linked to the average kinetic energy of molecules, gas pressure arises from wall collisions delivering impulse per unit area, and r.m.s. speed is the square root of mean square velocity.

Step-by-Step Solution

1

Identify the microscopic origin of temperature.

Temperature represents the average translational kinetic energy of gas molecules.

From the kinetic theory equation

\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T

, absolute temperature directly measures molecular kinetic energy.

2

Identify the microscopic origin of pressure.

Pressure is caused by molecular collisions with container walls.

Each collision transfers momentum to the wall; force is the time rate of momentum change, and force per unit area defines pressure.

3

Identify the definition of root-mean-square speed.

r.m.s. speed is the square root of the average of squared molecular speeds.

It accounts for the statistical distribution of molecular velocities in a gas sample.

Key Concept

Microscopic interpretation of macroscopic gas properties via Kinetic Theory of Matter

Question 7Question →

A sample of an ideal gas has an average translational kinetic energy of $E_k$ per molecule at an initial temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until the average kinetic energy per molecule doubles to $2E_k$ , what is the final temperature of the gas in degrees Celsius?

54^\circ\text{C}

327^\circ\text{C}

600^\circ\text{C}

927^\circ\text{C}

Show answer & explanation

Answer: $327^\circ\text{C}$

Answer

The final temperature of the gas is

327^\circ\text{C}

.

In the kinetic theory of matter, average translational kinetic energy per molecule is directly proportional to absolute temperature (

E_k \propto T

). Initial temperature

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

. Doubling the kinetic energy doubles the Kelvin temperature to

600\text{ K}

. Subtracting

273

yields

327^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

Kinetic theory calculations require thermodynamic temperature measured on the absolute Kelvin scale.

2

Apply the proportional relationship between average translational kinetic energy and temperature.

Since

E_k \propto T

, doubling

E_k

means

T_2 = 2 \times T_1 = 2 \times 300\text{ K} = 600\text{ K}

.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature.

3

Convert the final temperature from Kelvin back to degrees Celsius.

t_2 = 600\text{ K} - 273 = 327^\circ\text{C}

The question asks for the final temperature specifically in degrees Celsius.

Key Concept

Average translational kinetic energy of an ideal gas molecule is directly proportional to its absolute temperature (

E_k = \frac{3}{2} k_B T

).

Estimated Time:1m 30s

Question 8Question →

A gas sample enclosed in a rigid container of fixed volume has a root-mean-square (r.m.s.) speed of $500\text{ m s}^{-1}$ at a temperature of $127^\circ\text{C}$ . If the gas is heated until its pressure is quadrupled, what is the new r.m.s. speed of the gas molecules in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 1000

Answer

1000

According to kinetic theory, the pressure of a fixed volume of gas is directly proportional to its absolute temperature (

P \propto T

), and the root-mean-square speed of its molecules is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). Quadrupling the pressure quadruples the absolute temperature in Kelvin from

400\text{ K}

to

1600\text{ K}

. Since the speed scales as

\sqrt{4} = 2

, the initial r.m.s. speed of

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

doubles to

1000\text{ m s}^{-1}

.

Step-by-Step Solution

1

Convert the initial temperature to absolute temperature (Kelvin)

T_1 = 127^\circ\text{C} + 273 = 400\text{ K}

Kinetic theory relationships and gas laws require temperature in absolute units (Kelvin).

2

Determine the new absolute temperature based on the pressure change at constant volume

T_2 = 4 \times T_1 = 1600\text{ K}

For a fixed volume of gas, pressure is directly proportional to absolute temperature (

P \propto T

). Therefore, quadrupling the pressure quadruples the absolute temperature.

3

Calculate the new root-mean-square speed using the square root relationship

v_2 = v_1 \sqrt{\frac{T_2}{T_1}} = 500 \times \sqrt{4} = 1000\text{ m s}^{-1}

Root-mean-square speed is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Key Concept

Relationship between microscopic kinetic parameters (r.m.s. speed) and macroscopic state variables (pressure and absolute temperature)

Question 9Question →

Match each physical quantity or concept from the kinetic theory of gases on the left with its corresponding microscopic description or mathematical relation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Root-mean-square speed corresponds to

\sqrt{\frac{3kT}{m}}

; Average translational kinetic energy per molecule corresponds to

\frac{3}{2}kT

; Gas pressure corresponds to

\frac{1}{3}\rho v_{\text{rms}}^2

; and Absolute temperature corresponds to the macroscopic measure proportional to mean translational kinetic energy.

Each kinetic theory quantity correctly matches its corresponding microscopic formula and definition derived from fundamental assumptions of ideal gas particle behavior.

Step-by-Step Solution

1

Analyze the microscopic derivation of root-mean-square speed

From kinetic theory,

\bar{E}_k = \frac{1}{2}m v_{\text{rms}}^2 = \frac{3}{2}kT

, which yields

v_{\text{rms}} = \sqrt{\frac{3kT}{m}}

.

This establishes the relationship between molecular speed, temperature, and mass.

2

Identify the relationship for average translational kinetic energy per molecule

The average translational kinetic energy per molecule is given directly by

\bar{E}_k = \frac{3}{2}kT

.

The mean kinetic energy per degree of freedom is

\frac{1}{2}kT

, summing to

\frac{3}{2}kT

for three translational dimensions.

3

Relate macroscopic gas pressure to microscopic particle collisions

Gas pressure is expressed as

P = \frac{1}{3}\rho v_{\text{rms}}^2

based on continuous elastic collisions of gas molecules with the container walls.

Pressure represents the average force exerted per unit area by molecular collisions.

4

Define absolute temperature in terms of molecular kinetic energy

Absolute temperature

T

is the macroscopic physical property directly proportional to the average kinetic energy of the molecules.

This provides the thermodynamic definition of temperature from kinetic theory.

Key Concept

Microscopic properties of ideal gas molecules and kinetic derivation of pressure and temperature

Question 10Question →

A sample of gas enclosed in a vessel has a density of $0.90\text{ kg/m}^3$ and exerts a pressure of $3.0 \times 10^5\text{ N/m}^2$ on the walls of the vessel. Based on the kinetic theory of gases, what is the root-mean-square (r.m.s.) speed of the gas molecules in $\text{m/s}$ ?

Show answer & explanation

Answer: 1000

Answer

The root-mean-square speed of the gas molecules is

1000\text{ m/s}

.

By applying the kinetic theory formula

P = \frac{1}{3} \rho v_{\text{rms}}^2

, rearranging gives

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

. Substituting

P = 3.0 \times 10^5\text{ N/m}^2

and

\rho = 0.90\text{ kg/m}^3

results in

v_{\text{rms}} = \sqrt{\frac{9.0 \times 10^5}{0.90}} = \sqrt{1.0 \times 10^6} = 1000\text{ m/s}

.

Step-by-Step Solution

1

Identify the kinetic theory equation relating gas pressure, density, and microscopic molecular speed.

P = \frac{1}{3} \rho v_{\text{rms}}^2

According to the kinetic theory of gases, the macroscopic pressure exerted by gas molecules colliding with container walls is proportional to the gas density and the square of their r.m.s. speed.

2

Make

v_{\text{rms}}

the subject of the formula.

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

Multiplying both sides by

3

and dividing by density

\rho

isolates

v_{\text{rms}}^2

, taking the square root yields

v_{\text{rms}}

.

3

Substitute the given numerical values into the equation.

v_{\text{rms}} = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.90}} = \sqrt{1,000,000} = 1000\text{ m/s}

Performing the division yields

1.0 \times 10^6\text{ m}^2/\text{s}^2

, whose square root gives the speed in

\text{m/s}

.

Key Concept

Kinetic Theory Pressure Equation relating macroscopic pressure and density to microscopic root-mean-square velocity (

P = \frac{1}{3}\rho v_{\text{rms}}^2

).

Question 11Question →

Match each kinetic theory concept or microscopic property of an ideal gas on the left with its corresponding mathematical expression or derivation result on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair the momentum change per collision with

2 m v_x

, the single-molecule average wall force with

\frac{m v_x^2}{L}

, the kinetic energy density with

\frac{3}{2} P

, and the root-mean-square speed with

\sqrt{\frac{3 R T}{M}}

.

Each kinetic theory quantity is derived directly from fundamental principles of mechanics applied to gas particles. Elastic collision with a wall yields a momentum reversal of magnitude

2 m v_x

. Taking the round-trip collision frequency over length

L

yields an average force of

\frac{m v_x^2}{L}

. Linking microscopic kinetic energy density to pressure gives

\frac{E_k}{V} = \frac{3}{2} P

, and linking pressure to the ideal gas law for one mole yields

v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Step-by-Step Solution

1

Analyze momentum transfer during elastic collision of a molecule with a wall.

Initial momentum along the x-axis is

p_i = m v_x

and final momentum after elastic reflection is

p_f = -m v_x

. The change in momentum is

\Delta p_x = p_f - p_i = -2 m v_x

, which has a magnitude of

2 m v_x

.

Elastic collision conserves kinetic energy and reverses velocity direction perpendicular to the wall.

2

Calculate the time rate of momentum transfer to determine average force.

The round-trip distance between opposite walls separated by length

L

is

2L

, so the time between collisions with the same wall is

\Delta t = \frac{2L}{v_x}

. The average force is

F_x = \frac{\Delta p}{\Delta t} = \frac{2 m v_x}{2L / v_x} = \frac{m v_x^2}{L}

.

Newton's second law expresses force as the average rate of change of momentum.

3

Relate total translational kinetic energy density to gas pressure.

From kinetic theory, gas pressure is given by

P = \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2

. Since total kinetic energy

E_k = \frac{1}{2} N m v_{\text{rms}}^2

, we can express pressure as

P = \frac{2}{3} \left(\frac{E_k}{V}\right)

. Rearranging gives energy density

\frac{E_k}{V} = \frac{3}{2} P

.

Translational kinetic energy density is directly proportional to pressure with a factor of 3/2.

4

Derive the formula for root-mean-square velocity from macroscopic and microscopic gas equations.

Substitute density

\rho = \frac{M}{V}

(where

M

is molar mass) into

P = \frac{1}{3} \rho v_{\text{rms}}^2

, obtaining

P = \frac{M v_{\text{rms}}^2}{3 V}

. Since

P V = R T

for one mole of ideal gas,

R T = \frac{1}{3} M v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Connects microscopic speed distribution parameter with thermodynamic temperature and molar mass.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

Question 12Question →

According to the kinetic theory of gases, the pressure exerted by an ideal gas on the walls of its container is primarily caused by which of the following mechanisms?

The continuous elastic collisions of gas molecules with the walls of the container

The gravitational attraction pulling gas molecules toward the base of the container

Strong attractive intermolecular forces acting between adjacent gas particles

Inelastic collisions between gas molecules resulting in loss of total kinetic energy

Show answer & explanation

Answer: The continuous elastic collisions of gas molecules with the walls of the container

Answer

The pressure of an ideal gas is caused by the continuous elastic collisions of gas molecules with the container walls.

The correct answer states that pressure is caused by the continuous elastic collisions of gas molecules with the walls of the container. In kinetic theory, each wall collision imparts impulse to the container surface, yielding a net average force per unit area.

Step-by-Step Solution

1

Identify the microscopic origin of gas pressure in kinetic theory.

Gas molecules are in constant, random motion and frequently collide with the inner walls of the container.

Each collision results in a change of momentum of the gas molecule.

2

Relate molecular momentum change to force and pressure.

By Newton's second law, the rate of change of momentum produces a force on the wall, and force per unit area equals pressure.

Continuous elastic collisions create a steady macroscopic pressure on the walls.

Key Concept

Origin of Gas Pressure in Kinetic Theory

Estimated Time:45s

Question 13Question →

Match each kinetic theory concept or gas behavior statement on the left with its corresponding microscopic mechanism or physical condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Absolute temperature corresponds to the average translational kinetic energy of gas molecules; Pressure exerted by a gas corresponds to the rate of momentum transferred per unit area from elastic wall collisions; Direct physical evidence of continuous molecular motion corresponds to Brownian motion and diffusion; Conditions for real gas ideal behavior correspond to low pressure and high temperature.

Each kinetic theory concept correctly maps to its physical baseline: absolute temperature reflects average molecular translational kinetic energy; pressure results from momentum transfer during elastic collisions with walls; Brownian motion and diffusion provide direct physical evidence of random molecular motion; and real gases obey ideal gas behavior best under low pressure and high temperature conditions.

Step-by-Step Solution

1

Relate absolute temperature to microscopic particle properties.

Absolute temperature

T

is proportional to the average kinetic energy of translational motion of the gas molecules,

\bar{E}_k = \frac{3}{2}k_B T

.

Kinetic theory establishes temperature as a macroscopic measure of microscopic kinetic energy.

2

Identify the kinetic origin of gas pressure.

Molecules undergo elastic collisions with container walls, causing momentum change

\Delta p

per unit time, resulting in pressure

P = \frac{F}{A}

.

Macroscopic pressure is the cumulative force per unit area produced by constant particle impacts.

3

Determine experimental phenomena validating molecular motion.

Brownian motion (erratic motion of suspended pollen/smoke particles) and gas diffusion confirm molecular kinetic motion.

Unbalanced bombardment by invisible gas molecules causes visible random motion of suspended particles.

4

Establish validity conditions for ideal gas assumptions.

Real gases behave ideally at low pressures (large intermolecular distances make particle volume negligible) and high temperatures (high kinetic energy overcomes intermolecular attraction).

These conditions satisfy the fundamental postulates of the kinetic model of ideal gases.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

Question 14Question →

At a temperature of $27^\circ\text{C}$ , the root-mean-square (r.m.s.) speed of the molecules of an ideal gas is $300\text{ m/s}$ . What is the temperature of the gas, in degrees Celsius, when the r.m.s. speed of its molecules increases to $600\text{ m/s}$ ?

Show answer & explanation

Answer: 927

Answer

The temperature of the gas when the r.m.s. speed reaches

600\text{ m/s}

is

927^\circ\text{C}

.

According to kinetic theory, the root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). First convert the initial temperature to Kelvin:

27^\circ\text{C} + 273 = 300\text{ K}

. Since the speed doubles from

300\text{ m/s}

to

600\text{ m/s}

, the ratio of speeds is

2

, which means the absolute temperature ratio is

2^2 = 4

. Thus, the new absolute temperature is

4 \times 300\text{ K} = 1200\text{ K}

. Converting back to Celsius gives

1200 - 273 = 927^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas kinetic equations require absolute temperature in Kelvin.

2

Apply the proportional relationship between r.m.s. speed and absolute temperature

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

In the kinetic theory of gases, root-mean-square speed is directly proportional to the square root of absolute temperature.

3

Calculate the final absolute temperature

T_2

T_2 = 1200\text{ K}

Doubling the r.m.s. speed requires quadrupling the absolute temperature (

2^2 \times 300\text{ K} = 1200\text{ K}

).

4

Convert the calculated absolute temperature back to degrees Celsius

\theta_2 = 1200 - 273 = 927^\circ\text{C}

Subtract 273 from the Kelvin temperature to find the value in degrees Celsius.

Key Concept

Proportionality between root-mean-square speed and absolute temperature in kinetic theory of gases

Kinetic Theory of Matter and Pressure of Gases

Answer

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