Thermal Physics Practice Questions — JAMB UTME — Page 8

Question 141Question →

At an air temperature of $30^\circ\text{C}$ , the saturated vapour pressure of water is $32\text{ mmHg}$ . If the dew point of the atmosphere is $15^\circ\text{C}$ and the saturated vapour pressure of water at $15^\circ\text{C}$ is $12\text{ mmHg}$ , what is the relative humidity of the air?

37.5\%

50.0\%

62.5\%

75.0\%

Show answer & explanation

Answer: $37.5\%$

Answer

The relative humidity of the air is

37.5\%

.

The relative humidity is defined as the ratio of the saturated vapour pressure at the dew point temperature to the saturated vapour pressure at the current air temperature. Substituting the given values gives

\frac{12\text{ mmHg}}{32\text{ mmHg}} \times 100\% = 37.5\%

.

Step-by-Step Solution

1

Identify the formula for relative humidity in terms of vapour pressures

\text{Relative Humidity (R.H.)} = \frac{\text{Partial Vapour Pressure of Water at Air Temp}}{\text{Saturated Vapour Pressure at Air Temp}} \times 100\%

By definition, the actual partial vapour pressure of water in unsaturated air equals the saturated vapour pressure at its dew point temperature.

2

Substitute the given values into the equation

\text{R.H.} = \frac{12\text{ mmHg}}{32\text{ mmHg}} \times 100\%

The SVP at dew point (

15^\circ\text{C}

) is

12\text{ mmHg}

and the SVP at air temperature (

30^\circ\text{C}

) is

32\text{ mmHg}

.

3

Calculate the percentage

\text{R.H.} = 0.375 \times 100\% = 37.5\%

Simplifying

\frac{12}{32}

gives

\frac{3}{8}

, which equals

0.375

or

37.5\%

.

Key Concept

Relative humidity is the ratio of the actual vapour pressure present in the air (saturated vapour pressure at the dew point) to the maximum saturated vapour pressure the air can hold at its current temperature, expressed as a percentage.

Question 142Question →

Match each practical thermal design feature on the left with its corresponding primary heat transfer mechanism or mitigation strategy on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Silvered surfaces match radiation reduction by reflection; Evacuated space matches elimination of conduction and convection via vacuum; Heating element at the bottom matches maximization of convection currents; Blackened copper cooling fins match heat dissipation via high conduction and radiation emissivity.

Each feature correctly pairs with its primary physical heat transfer process: silvered surfaces reflect radiant waves; vacuum stops conduction and convection due to lack of matter; bottom heating drives convection loops via fluid expansion; and blackened copper fins combine high thermal conduction with maximum radiative heat emission.

Step-by-Step Solution

1

Analyze the silvered inner surfaces of a vacuum flask.

Silver acts as a polished mirror to electromagnetic thermal radiation.

Bright, shiny surfaces reflect radiation instead of absorbing or emitting it.

2

Analyze the evacuated space (vacuum) between the walls.

Conduction (requiring atomic vibrations/collisions) and convection (requiring bulk fluid movement) cannot occur.

A vacuum lacks any material particles required to conduct heat or sustain convective fluid flows.

3

Analyze the bottom placement of a kettle's heating element.

Heated liquid expands, decreases in density, and rises while denser cold liquid falls.

Placing the element at the bottom establishes upward thermal convection loops throughout the entire volume.

4

Analyze blackened copper cooling fins on refrigerator coils.

Copper rapidly conducts heat away from pipes, and the black coating maximizes infrared radiation into the surrounding air.

Dark, dull surfaces have high thermal emissivity for radiation, while metals like copper have high thermal conductivity.

Key Concept

Modes of Heat Transfer (Conduction, Convection, and Radiation)

Estimated Time:1m 30s

Question 143Question →

Match each physical scenario on the left with its corresponding primary mechanism of heat transfer on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warm air circulation corresponds to Convection; Heat transfer along a solid aluminum rod corresponds to Conduction; Energy transfer across an evacuated space corresponds to Radiation.

Each scenario uniquely exemplifies a fundamental mode of heat transfer: rising warmer air in a fluid is driven by buoyant convection currents; heat flow through a solid rod is driven by conduction via lattice vibrations and electron drift; energy transport across a vacuum is mediated exclusively by electromagnetic radiation.

Step-by-Step Solution

1

Analyze the mechanism behind warm air rising and cool air sinking above a heater.

Heating causes fluid expansion, lowering its density. The buoyant force causes warm fluid to rise while cooler, denser fluid sinks, forming convection currents.

Convection is the bulk movement of fluid particles driven by temperature-induced density gradients.

2

Analyze the mechanism of thermal transfer along a solid aluminum rod.

Heat flows through the solid medium from particle to particle via atomic vibrations and free electron motion without bulk displacement of the aluminum.

Conduction is the mode of heat transfer in solids mediated by molecular collisions and free electron diffusion.

3

Analyze the energy transfer through an evacuated glass chamber.

Because there is no material medium inside the vacuum, neither conduction nor convection can occur. Heat travels as infrared electromagnetic waves.

Radiation is the only mode of heat transfer that can propagate through a vacuum.

Key Concept

Distinguishing mechanisms of heat transfer: Conduction (solids), Convection (fluids), and Radiation (vacuum/electromagnetic waves)

Question 144Question →

When equal masses of two different substances absorb equal quantities of heat energy without undergoing a phase change, the substance with the lower specific heat capacity will experience a larger temperature rise.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because temperature change is inversely proportional to specific heat capacity when equal heat energy is supplied to equal masses.

According to the heat equation

Q = m c \Delta T

, the temperature change is given by

\Delta T = \frac{Q}{m c}

. For identical values of mass

m

and heat input

Q

, temperature change

\Delta T

is inversely proportional to specific heat capacity

c

. Therefore, a substance with a lower specific heat capacity will undergo a larger temperature rise.

Step-by-Step Solution

1

Identify the formula relating heat energy absorbed to mass, specific heat capacity, and temperature change.

The relationship is

Q = m c \Delta T

.

This fundamental equation models sensible heat exchange in matter.

2

Isolate the temperature change variable

\Delta T

.

\Delta T = \frac{Q}{m c}.

Expressing

\Delta T

explicitly allows direct analysis of how specific heat capacity influences temperature rise.

3

Determine the functional dependence of

\Delta T

on

c

under constant

m

and

Q

.

\Delta T \propto \frac{1}{c}.

Since

\Delta T

and

c

are inversely proportional, a smaller specific heat capacity

c

leads to a larger temperature increase

\Delta T

.

Key Concept

Inverse proportionality between specific heat capacity and temperature change for a given mass and heat input.

Question 145Question →

The saturated vapour pressure of water in an enclosed space is $25\text{ mmHg}$ . If the relative humidity within the space is measured to be $64\%$ , what is the partial pressure of the water vapour in $\text{mmHg}$ ?

Show answer & explanation

Answer: 16

Answer

The partial pressure of the water vapour in the enclosed space is

16\text{ mmHg}

.

Relative humidity is defined as the ratio of the actual partial pressure of water vapour present in a given volume of air to the saturated vapour pressure at the same temperature, expressed as a percentage:

\text{R.H.} = \frac{P}{\text{S.V.P.}} \times 100\%

. Rearranging this equation gives

P = \frac{\text{R.H.} \times \text{S.V.P.}}{100} = \frac{64 \times 25}{100} = 16\text{ mmHg}

.

Step-by-Step Solution

1

Identify the given parameters and state the relative humidity formula.

Relative Humidity (R.H.) =

64\%

, Saturated Vapour Pressure (S.V.P.) =

25\text{ mmHg}

.

\text{R.H.} = \frac{P}{\text{S.V.P.}} \times 100\%

.

Relative humidity relates the actual partial vapour pressure present to the maximum saturated vapour pressure possible at that temperature.

2

Substitute the values into the equation and solve for partial pressure

P

.

64 = \frac{P}{25} \times 100 \implies 64 = 4P \implies P = 16\text{ mmHg}

.

Dividing

100

by

25

yields a factor of

4

, allowing clean mental computation.

Key Concept

Relationship between relative humidity, partial vapour pressure, and saturated vapour pressure.

Question 146Question →

An electric heater rated at $200\text{ W}$ is used to heat a liquid of mass $1.5\text{ kg}$ for $7\text{ minutes}$ . If the temperature of the liquid rises from $25^\circ\text{C}$ to $65^\circ\text{C}$ , assuming no heat energy is lost to the surroundings, what is the specific heat capacity of the liquid in $\text{J}\text{kg}^{-1}\text{K}^{-1}$ ?

Show answer & explanation

Answer: 1400

Answer

The specific heat capacity of the liquid is

1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

The electrical energy supplied over

7\text{ minutes}

(

420\text{ s}

) at a power rating of

200\text{ W}

equals

84,000\text{ J}

. Equating this total energy to thermal absorption

Q = m c \Delta T

for a

1.5\text{ kg}

mass experiencing a

40\text{ K}

temperature rise gives

c = \frac{84,000}{1.5 \times 40} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

Step-by-Step Solution

1

Convert heating time to seconds and compute total heat energy supplied.

t = 7 \times 60 = 420\text{ s}

, so

Q = P \times t = 200 \times 420 = 84,000\text{ J}

.

Electrical power is the rate of energy transfer (

P = Q/t

), so heat energy equals power multiplied by time in seconds.

2

Calculate temperature difference.

\Delta T = 65^\circ\text{C} - 25^\circ\text{C} = 40\text{ K}

.

The temperature rise drives heat absorption according to the thermal equation.

3

Rearrange the heat energy formula

Q = m c \Delta T

to find the specific heat capacity

c

.

c = \frac{84,000}{1.5 \times 40} = \frac{84,000}{60} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

.

Specific heat capacity represents the heat energy required per unit mass per unit temperature change.

Key Concept

Specific heat capacity calculation using electrical energy input (

Q = P t = m c \Delta T

).

Estimated Time:1m 15s

Question 147Question →

A mercury barometer reads an atmospheric pressure of $760\text{ mmHg}$ . A small quantity of volatile liquid is introduced into the space above the mercury column, where it saturates the space with vapour. If the saturated vapour pressure of the liquid at that temperature is $40\text{ mmHg}$ , what is the new height of the mercury column in the barometer tube?

720\text{ mmHg}

800\text{ mmHg}

760\text{ mmHg}

40\text{ mmHg}

Show answer & explanation

Answer: $720\text{ mmHg}$

Answer

720\text{ mmHg}

Atmospheric pressure supports both the height of the liquid mercury column and the downward pressure exerted by the saturated vapour. Therefore, the height of the mercury column is given by subtracting the saturated vapour pressure (

40\text{ mmHg}

) from atmospheric pressure (

760\text{ mmHg}

), yielding

720\text{ mmHg}

.

Step-by-Step Solution

1

Identify the equilibrium condition for atmospheric pressure balancing the barometer contents.

Atmospheric pressure equals the sum of the mercury column pressure and the saturated vapour pressure:

P_{\text{atm}} = P_{\text{Hg}} + P_{\text{SVP}}

.

The trapped vapour exerts a downward force on top of the liquid mercury column inside the sealed tube.

2

Substitute the given values into the pressure equation and solve for the mercury height.

760\text{ mmHg} = P_{\text{Hg}} + 40\text{ mmHg} \implies P_{\text{Hg}} = 760\text{ mmHg} - 40\text{ mmHg} = 720\text{ mmHg}

.

Subtracting the saturated vapour pressure from atmospheric pressure gives the net height of mercury the atmosphere can support.

Key Concept

Effect of Saturated Vapour Pressure on Barometric Height

Question 148Question →

During the vaporization of a pure liquid at its boiling point under constant atmospheric pressure, the absorbed thermal energy increases the average kinetic energy of the molecules while maintaining a constant temperature.

True

False

Show answer & explanation

Answer: False

Answer

False. Temperature is directly proportional to the average kinetic energy of molecules. Because temperature remains constant during phase changes, the average kinetic energy remains unchanged, while the latent heat increases molecular potential energy.

The correct answer is False because temperature measures average kinetic energy directly. Since temperature remains constant during boiling, molecular kinetic energy cannot increase; instead, heat input goes entirely into increasing molecular potential energy.

Step-by-Step Solution

1

Relate temperature to molecular kinetic energy.

By the kinetic theory of matter, absolute temperature

T

is directly proportional to the average translational kinetic energy

\frac{1}{2}m\bar{v}^2

of the molecules.

Temperature is a microscopic measure of molecular motion.

2

Examine the state of temperature during vaporization at the boiling point.

During boiling, temperature remains constant until the phase change from liquid to gas is completely finished.

Latent heat absorption occurs at constant temperature.

3

Deduce which molecular energy component changes during latent heat absorption.

Since temperature does not change, average kinetic energy stays constant. The absorbed latent heat increases intermolecular potential energy by breaking intermolecular bonds and performing work during expansion against atmospheric pressure.

Thermal energy supplied during a phase transition goes into potential energy rather than kinetic energy.

Key Concept

Microscopic interpretation of latent heat and temperature during a phase change

Question 149Question →

An electric heater rated at $200\text{ W}$ is used to heat a $0.40\text{ kg}$ sample of a pure solid substance maintained at its melting point. If it takes $2\text{ minutes}$ for exactly half of the sample to melt into liquid at the same temperature, what is the specific latent heat of fusion of the substance?

1.2 \times 10^5\text{ J kg}^{-1}

6.0 \times 10^4\text{ J kg}^{-1}

2.0 \times 10^3\text{ J kg}^{-1}

1.2 \times 10^7\text{ J kg}^{-1}

Show answer & explanation

Answer: $1.2 \times 10^5\text{ J kg}^{-1}$

Answer

The specific latent heat of fusion of the substance is

1.2 \times 10^5\text{ J kg}^{-1}

.

The thermal energy absorbed during melting is

Q = P \times t = 200\text{ W} \times 120\text{ s} = 24,000\text{ J}

. Since only half of the

0.40\text{ kg}

sample melts, the mass undergoing phase change is

0.20\text{ kg}

. Dividing the heat absorbed (

24,000\text{ J}

) by the melted mass (

0.20\text{ kg}

) yields the correct specific latent heat of fusion of

1.2 \times 10^5\text{ J kg}^{-1}

.

Step-by-Step Solution

1

Calculate total thermal energy supplied by the heater.

Q = P \times t = 200\text{ W} \times (2 \times 60\text{ s}) = 24,000\text{ J}

Electrical energy supplied is converted completely into thermal energy.

2

Determine the mass of substance that melted.

m_{\text{melted}} = \frac{1}{2} \times 0.40\text{ kg} = 0.20\text{ kg}

Only half of the total mass undergoes phase change.

3

Compute the specific latent heat of fusion

L_f

.

L_f = \frac{Q}{m_{\text{melted}}} = \frac{24,000\text{ J}}{0.20\text{ kg}} = 120,000\text{ J kg}^{-1} = 1.2 \times 10^5\text{ J kg}^{-1}

Phase change occurs at constant temperature using the latent heat equation

Q = m L_f

.

Key Concept

Latent Heat of Fusion and Energy Balance

Question 150Question →

A solid metal component of mass $0.6\text{ kg}$ absorbs $5400\text{ J}$ of thermal energy when its temperature increases from $30^\circ\text{C}$ to $75^\circ\text{C}$ . Calculate the heat capacity of the component in $\text{J K}^{-1}$ .

Show answer & explanation

Answer: 120

Answer

The heat capacity of the metal component is

120\text{ J K}^{-1}

.

Heat capacity (

C

) is calculated by dividing the total thermal energy absorbed (

Q = 5400\text{ J}

) by the resulting temperature change (

\Delta T = 75^\circ\text{C} - 30^\circ\text{C} = 45\text{ K}

). Performing the division yields

C = \frac{5400}{45} = 120\text{ J K}^{-1}

. Note that the mass (

0.6\text{ kg}

) is not included in the denominator because heat capacity refers to the entire object, unlike specific heat capacity.

Step-by-Step Solution

1

Determine the change in temperature

\Delta T = 75^\circ\text{C} - 30^\circ\text{C} = 45\text{ K}

Heat capacity depends on the temperature difference through which heat energy is absorbed.

2

Apply the definition of heat capacity

C = \frac{Q}{\Delta T}

Heat capacity (

C

) is defined as the thermal energy required to raise the temperature of the entire body by

1\text{ K}

(or

1^\circ\text{C}

).

3

Substitute the given thermal energy and temperature change

C = \frac{5400\text{ J}}{45\text{ K}} = 120\text{ J K}^{-1}

Dividing total energy by total temperature rise gives heat capacity.

Key Concept

Heat capacity (

C

) represents the thermal energy needed to change a body's temperature by one unit (

C = Q / \Delta T

), whereas specific heat capacity (

c

) is heat capacity per unit mass (

c = C / m

).

Question 151Question →

A thermistor has an electrical resistance of $800\,\Omega$ at the melting point of ice ( $0^\circ\text{C}$ ) and $200\,\Omega$ at the boiling point of water ( $100^\circ\text{C}$ ). Assuming the thermometric property varies linearly with temperature, what is the temperature in degrees Celsius ( $^\circ\text{C}$ ) when the resistance of the thermistor is $500\,\Omega$ ?

Show answer & explanation

Answer: 50

Answer

The temperature corresponding to a resistance of

500\,\Omega

is

50^\circ\text{C}

.

Applying the standard thermometric interpolation relation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

with

R_0 = 800\,\Omega

,

R_{100} = 200\,\Omega

, and

R_\theta = 500\,\Omega

gives

\theta = \frac{500 - 800}{200 - 800} \times 100 = \frac{-300}{-600} \times 100 = 50^\circ\text{C}

.

Step-by-Step Solution

1

Identify the values of the thermometric property at the ice point and steam point.

R_0 = 800\,\Omega

and

R_{100} = 200\,\Omega

.

These established values represent the fixed points of

0^\circ\text{C}

and

100^\circ\text{C}

respectively.

2

Set up the linear relationship formula for temperature conversion on the Celsius scale.

$\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}$

The temperature scale is defined linearly between the two fixed calibration points.

3

Substitute the unknown resistance

R_\theta = 500\,\Omega

into the equation and compute the result.

$\theta = \frac{500 - 800}{200 - 800} \times 100^\circ\text{C} = \frac{-300}{-600} \times 100^\circ\text{C} = 50^\circ\text{C}$

Dividing the change from the lower fixed point by the total interval between fixed points gives the fraction of

100^\circ\text{C}

.

Key Concept

Linear relationship between a thermometric property and temperature

Question 152Question →

A metal container with a volume of $200\text{ cm}^3$ at $30^\circ\text{C}$ is completely filled with an oil. The linear expansivity of the metal container is $2.0 \times 10^{-5}\text{ K}^{-1}$ and the real cubic expansivity of the oil is $6.0 \times 10^{-4}\text{ K}^{-1}$ . If the temperature of the system is raised to $80^\circ\text{C}$ , what volume of oil (in $\text{cm}^3$ ) will overflow?

Show answer & explanation

Answer: 5.4

Answer

The volume of oil that overflows is

5.4\text{ cm}^3

.

When a container filled to the brim with liquid is heated, both the liquid and container expand. The apparent cubic expansivity of the liquid is the difference between its real cubic expansivity and the volumetric expansivity of the vessel (3 * alpha). Subtracting 0.6 * 10^-4 K^-1 from 6.0 * 10^-4 K^-1 gives an apparent cubic expansivity of 5.4 * 10^-4 K^-1. Multiplying this by the initial volume of 200 cm^3 and the temperature rise of 50 K gives an overflow volume of 5.4 cm^3.

Step-by-Step Solution

1

Determine the temperature increase (\Delta T)

\Delta T = 80^\circ\text{C} - 30^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Calculate the cubic expansivity of the container (\gamma_v)

\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

The volumetric expansivity of an isotropic solid container is three times its linear expansivity.

3

Calculate the apparent cubic expansivity of the liquid (\gamma_a)

\gamma_a = \gamma_r - \gamma_v = 6.0 \times 10^{-4}\text{ K}^{-1} - 0.6 \times 10^{-4}\text{ K}^{-1} = 5.4 \times 10^{-4}\text{ K}^{-1}

Apparent expansivity accounts for the simultaneous expansion of the vessel containing the liquid.

4

Calculate the volume of liquid that overflows (\Delta V)

\Delta V = V_0 \times \gamma_a \times \Delta T = 200\text{ cm}^3 \times 5.4 \times 10^{-4}\text{ K}^{-1} \times 50\text{ K} = 5.4\text{ cm}^3

The overflow volume equals the apparent expansion of the liquid volume.

Key Concept

Real and Apparent Expansion of Liquids

Question 153Question →

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 154Question →

A high-altitude weather research probe contains $0.030\text{ m}^3$ of helium gas at a pressure of $1.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ at ground level. During ascent, the gas expands to a volume of $0.050\text{ m}^3$ while its temperature rises to $127^\circ\text{C}$ . Assuming the helium behaves as an ideal gas, what is the new pressure of the gas in the probe?

4.50 \times 10^4\text{ Pa}

8.00 \times 10^4\text{ Pa}

2.22 \times 10^5\text{ Pa}

2.82 \times 10^5\text{ Pa}

Show answer & explanation

Answer: $8.00 \times 10^4\text{ Pa}$

Answer

The new pressure of the helium gas is

8.00 \times 10^4\text{ Pa}

.

Converting temperatures to absolute scale gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Rearranging the combined gas equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

for final pressure

P_2

yields

P_2 = 1.00 \times 10^5 \times \frac{0.030}{0.050} \times \frac{400}{300} = 8.00 \times 10^4\text{ Pa}

.

Step-by-Step Solution

1

Convert temperatures from Celsius to Kelvin

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

and

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

Gas laws require absolute temperature in Kelvin to maintain proportional thermodynamic relationships.

2

Apply the combined gas law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}

Mass of the gas is constant while pressure, volume, and temperature all change simultaneously.

3

Substitute given values and calculate final pressure

P_2 = 1.00 \times 10^5 \times \frac{0.030}{0.050} \times \frac{400}{300} = 8.00 \times 10^4\text{ Pa}

Multiplying initial pressure by the volume contraction ratio (

0.60

) and temperature expansion ratio (

1.333

) gives the new pressure.

Key Concept

Combined Gas Law

Question 155Question →

A glass vessel has a linear expansivity of $9.0 \times 10^{-6}\text{ K}^{-1}$ . When filled to the brim with $500\text{ cm}^3$ of a liquid at $20^\circ\text{C}$ and heated to $70^\circ\text{C}$ , $12\text{ cm}^3$ of the liquid overflows. What is the real cubic expansivity of the liquid?

4.80 \times 10^{-4}\text{ K}^{-1}

4.89 \times 10^{-4}\text{ K}^{-1}

5.07 \times 10^{-4}\text{ K}^{-1}

4.53 \times 10^{-4}\text{ K}^{-1}

Show answer & explanation

Answer: $5.07 \times 10^{-4}\text{ K}^{-1}$

Answer

The real cubic expansivity of the liquid is

5.07 \times 10^{-4}\text{ K}^{-1}

.

The real cubic expansivity of a liquid accounts for both the liquid's apparent expansion (overflow volume) and the expansion of the containing vessel. Calculating the apparent expansivity gives

\gamma_a = 12 / (500 \times 50) = 4.80 \times 10^{-4}\text{ K}^{-1}

. Combining this with the vessel's cubical expansivity

\gamma_v = 3 \times 9.0 \times 10^{-6} = 2.70 \times 10^{-5}\text{ K}^{-1}

yields

\gamma_r = 4.80 \times 10^{-4} + 0.27 \times 10^{-4} = 5.07 \times 10^{-4}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

) and the apparent cubic expansivity (

\gamma_a

).

\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50\text{ K}

.

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{12\text{ cm}^3}{500\text{ cm}^3 \times 50\text{ K}} = 4.80 \times 10^{-4}\text{ K}^{-1}

.

Apparent expansion corresponds directly to the volume of liquid that overflows.

2

Calculate the cubic expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3 \times \alpha_v = 3 \times (9.0 \times 10^{-6}\text{ K}^{-1}) = 2.70 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}

.

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

3

Calculate the real cubic expansivity of the liquid (

\gamma_r

).

\gamma_r = \gamma_a + \gamma_v = 4.80 \times 10^{-4}\text{ K}^{-1} + 0.27 \times 10^{-4}\text{ K}^{-1} = 5.07 \times 10^{-4}\text{ K}^{-1}

.

Real cubic expansivity accounts for both the apparent expansion of the liquid and the volume expansion of the container.

Key Concept

Relationship between real cubic expansivity, apparent cubic expansivity, and container volume expansion:

\gamma_r = \gamma_a + \gamma_v

.

Question 156Question →

A rigid gas vessel sealed on a meteorological satellite initially contains a fixed volume of an ideal gas at a pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . During an orbital maneuver, solar radiation heats the vessel to $327^\circ\text{C}$ , while a small vent valve simultaneously allows $20\%$ of the gas molecules to escape. Assuming the volume of the vessel remains constant, what is the final pressure of the gas inside the vessel?

2.40 \times 10^5\text{ Pa}

1.45 \times 10^6\text{ Pa}

3.00 \times 10^5\text{ Pa}

6.00 \times 10^4\text{ Pa}

Show answer & explanation

Answer: $2.40 \times 10^5\text{ Pa}$

Answer

2.40 \times 10^5\text{ Pa}

According to the ideal gas law (

PV = nRT

), when volume is held constant, pressure is directly proportional to both the number of moles of gas present and the absolute temperature in Kelvin (

P \propto nT

). First, convert the initial and final temperatures to Kelvin:

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

and

T_2 = 327 + 273 = 600\text{ K}

. Because

20\%

of the gas escapes,

80\%

remains inside the vessel, so

n_2 = 0.80n_1

. Calculating final pressure yields

P_2 = P_1 \times (n_2/n_1) \times (T_2/T_1) = 1.50 \times 10^5 \times 0.80 \times (600/300) = 2.40 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert all given temperatures from degrees Celsius to absolute temperature in Kelvin.

T_1 = 27 + 273.15 = 300\text{ K}

,

T_2 = 327 + 273.15 = 600\text{ K}

Gas laws and the ideal gas equation require absolute temperature in Kelvin.

2

Determine the fraction of gas remaining in the rigid vessel.

If

20\%

escapes, remaining fraction

n_2 = (1 - 0.20)n_1 = 0.80n_1

Pressure depends on the quantity of gas remaining inside the fixed volume.

3

Apply the ideal gas equation

PV = nRT

for constant volume

V

.

P_2 = P_1 \times \left(\frac{n_2}{n_1}\right) \times \left(\frac{T_2}{T_1}\right)

Since volume is constant, pressure is directly proportional to the product of amount of gas and absolute temperature.

4

Substitute known values to compute the final pressure

P_2

.

P_2 = (1.50 \times 10^5\text{ Pa}) \times 0.80 \times \left(\frac{600\text{ K}}{300\text{ K}}\right) = 2.40 \times 10^5\text{ Pa}

Multiplying the initial pressure by the mole fraction ratio and temperature ratio gives the final gas pressure.

Key Concept

Ideal Gas Equation and Combined Gas Behavior

Estimated Time:2m 0s

Question 157Question →

A sealed rigid glass bulb contains a fixed mass of helium gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Assuming the volume of the bulb remains constant, calculate the final pressure of the gas in $\text{Pa}$ when it is heated to $127^\circ\text{C}$ .

Show answer & explanation

Answer: 160000

Answer

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

(

160\,000\text{ Pa}

).

The correct result of

160\,000\text{ Pa}

(

1.60 \times 10^5\text{ Pa}

) is obtained by applying the Pressure Law

\frac{P_1}{T_1} = \frac{P_2}{T_2}

using absolute temperatures (

T_1 = 300\text{ K}

,

T_2 = 400\text{ K}

).

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to the absolute temperature scale (Kelvin).

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

and

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

.

All gas law equations require absolute temperatures in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) for a fixed mass of gas at constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

The pressure of a gas is directly proportional to its absolute temperature when volume remains constant.

3

Substitute the known values to find

P_2

.

P_2 = 1.20 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa} = 160\,000\text{ Pa}

.

Multiplying

1.20 \times 10^5

by the temperature ratio

\frac{4}{3}

gives

1.60 \times 10^5\text{ Pa}

.

Key Concept

Pressure Law (Gay-Lussac's Law) and absolute temperature conversion

Question 158Question →

A rigid container $A$ of volume $0.060\text{ m}^3$ contains an ideal gas at an absolute pressure of $4.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Container $A$ is connected via a narrow tube of negligible volume with a valve to an evacuated rigid container $B$ of volume $0.040\text{ m}^3$ . The valve is opened, allowing gas to flow between the containers until equilibrium is established. If container $A$ is maintained at $27^\circ\text{C}$ while container $B$ is maintained at $127^\circ\text{C}$ , what is the final equilibrium pressure of the gas in kilopascals (kPa)?

Show answer & explanation

Answer: 300

Answer

300 kPa

Converting both temperatures to Kelvin (

300\text{ K}

and

400\text{ K}

) and applying mole conservation (

n_{\text{total}} = n_A + n_B

) yields a final uniform pressure of

3.00 \times 10^5\text{ Pa}

, which equals

300\text{ kPa}

.

Step-by-Step Solution

1

Convert temperatures from degrees Celsius to kelvins.

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

and

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

.

Gas laws and calculations using the ideal gas equation require absolute temperatures in Kelvin.

2

Calculate the initial number of moles of gas present in the system.

n_{\text{total}} = \frac{p_A V_A}{R T_A} = \frac{(4.50 \times 10^5\text{ Pa})(0.060\text{ m}^3)}{R(300\text{ K})} = \frac{90}{R}

.

Container B is initially evacuated, meaning all gas molecules originate from container A.

3

Express the total number of moles at final equilibrium in terms of final pressure

P

.

n_{\text{final}} = n_A + n_B = \frac{P V_A}{R T_A} + \frac{P V_B}{R T_B} = \frac{P}{R} \left(\frac{0.060}{300} + \frac{0.040}{400}\right) = \frac{P}{R} (2.0 \times 10^{-4} + 1.0 \times 10^{-4}) = \frac{3.0 \times 10^{-4} P}{R}

.

At equilibrium, the pressure

P

becomes uniform throughout both interconnected containers.

4

Equate the initial and final total moles to solve for the equilibrium pressure

P

.

\frac{90}{R} = \frac{3.0 \times 10^{-4} P}{R} \implies P = \frac{90}{3.0 \times 10^{-4}} = 3.00 \times 10^5\text{ Pa} = 300\text{ kPa}

.

The total mass and number of moles of gas are conserved within the sealed system.

Key Concept

Conservation of total moles in interconnected gas containers governed by the ideal gas equation

PV = nRT

Estimated Time:2m 30s

Question 159Question →

An air bubble with an initial volume of $4.0\text{ cm}^3$ is released by a scuba diver at a depth where the total pressure is $2.50 \times 10^5\text{ Pa}$ and the water temperature is $7^\circ\text{C}$ . Calculate the volume of the bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the water temperature is $77^\circ\text{C}$ .

Show answer & explanation

Answer: 12.5

Answer

12.5 cm³

Converting temperatures to the Kelvin scale (

T_1 = 280\text{ K}

and

T_2 = 350\text{ K}

) and applying the combined gas law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

gives a final volume of

12.5\text{ cm}^3

.

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to absolute temperatures in Kelvin.

T_1 = 7^\circ\text{C} + 273 = 280\text{ K}

and

T_2 = 77^\circ\text{C} + 273 = 350\text{ K}

.

Gas law calculations require thermodynamic temperature measured on the Kelvin scale.

2

Set up the combined gas law relationship for a fixed mass of gas.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The mass of air inside the bubble remains constant while pressure, volume, and temperature change simultaneously.

3

Substitute the known values into the equation and solve for the final volume

V_2

.

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{(2.50 \times 10^5\text{ Pa}) \times (4.0\text{ cm}^3) \times (350\text{ K})}{(1.00 \times 10^5\text{ Pa}) \times (280\text{ K})} = 12.5\text{ cm}^3

.

Algebraic substitution yields the correct final volume of the expanded bubble.

Key Concept

Combined Gas Law

Question 160Question →

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

Thermal Physics

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