Thermal Physics Practice Questions — JAMB UTME — Page 3

Question 41Question →

A heavy-duty truck tire contains a fixed mass of air at an initial absolute pressure of $2.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . After traveling a long distance, friction causes the temperature of the air inside the tire to increase to $57^\circ\text{C}$ while its volume remains constant. What is the new absolute pressure of the air inside the tire in pascals ( $\text{Pa}$ )?

Show answer & explanation

Answer: 220000

Answer

The new absolute pressure of the air inside the tire is

220,000\text{ Pa}

(or

2.20 \times 10^5\text{ Pa}

).

According to Gay-Lussac's Law, at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 330\text{ K}

. Calculating

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

gives

2.00 \times 10^5\text{ Pa} \times \frac{330}{300} = 220,000\text{ Pa}

.

Step-by-Step Solution

1

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

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

and

T_2 = 57^\circ\text{C} + 273 = 330\text{ K}

.

All gas law equations require absolute temperatures in Kelvin.

2

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

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

When volume is constant, gas pressure is directly proportional to absolute temperature.

3

Substitute the known values into the equation to calculate the final pressure

P_2

.

P_2 = 2.00 \times 10^5\text{ Pa} \times \frac{330\text{ K}}{300\text{ K}} = 2.20 \times 10^5\text{ Pa} = 220,000\text{ Pa}

.

Multiplying the initial pressure by the ratio of absolute temperatures gives the new pressure.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 42Question →

A flexible research balloon is filled with $1.50\text{ m}^3$ of helium gas at a temperature of $27^\circ\text{C}$ . If the gas is heated at constant pressure until its temperature reaches $127^\circ\text{C}$ , what is the new volume of the balloon?

2.00\text{ m}^3

7.06\text{ m}^3

1.13\text{ m}^3

1.83\text{ m}^3

Show answer & explanation

Answer: $2.00\text{ m}^3$

Answer

The new volume of the balloon is

2.00\text{ m}^3

.

The answer of

2.00\text{ m}^3

correctly uses Charles's Law with absolute temperatures converted to Kelvin (

300\text{ K}

and

400\text{ K}

), showing that heating the gas causes a proportional volume expansion.

Step-by-Step Solution

1

Convert temperatures from Celsius to the Kelvin absolute scale.

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

,

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

.

Gas laws strictly require thermodynamic (absolute) temperature measured in Kelvin.

2

Apply Charles's Law for constant pressure processes.

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

.

Volume is directly proportional to absolute temperature when pressure remains constant.

3

Substitute the known values and calculate

V_2

.

V_2 = 1.50\text{ m}^3 \times \frac{400\text{ K}}{300\text{ K}} = 2.00\text{ m}^3

.

Simplifying the fraction

\frac{400}{300} = \frac{4}{3}

gives

1.50 \times \frac{4}{3} = 2.00\text{ m}^3

.

Key Concept

Charles's Law (

V_1/T_1 = V_2/T_2

at constant pressure)

Estimated Time:1m 15s

Question 43Question →

When a pure substance undergoes a change of state from liquid to vapor at its boiling point under constant atmospheric pressure, the added thermal energy increases the average kinetic energy of the molecules, causing its temperature to rise continuously until vaporization is complete.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. During a phase change at boiling point, thermal energy acts as latent heat, increasing the potential energy of the molecules to break intermolecular bonds while keeping average kinetic energy and temperature constant.

During boiling under constant atmospheric pressure, all absorbed thermal energy is utilized as latent heat of vaporization to do work against intermolecular forces, increasing molecular potential energy. Because temperature measures average kinetic energy, the temperature remains strictly constant until the phase change is complete.

Step-by-Step Solution

1

Identify the nature of heat added during a state change at constant pressure.

The added heat is latent heat of vaporization.

Heat added during a change of phase at boiling point does not produce a temperature change.

2

Relate molecular kinetic energy to temperature.

Temperature is directly proportional to the average translational kinetic energy of the molecules.

If kinetic energy increases, temperature must rise; since temperature remains constant during boiling, average kinetic energy does not change.

3

Determine the destination of the absorbed thermal energy.

Energy is stored as molecular potential energy.

The energy is spent overcoming attractive intermolecular forces to separate molecules from the liquid state into the gaseous state.

Key Concept

Constant temperature and molecular energy changes during phase transition

Question 44Question →

A container of heat capacity $80\text{ J K}^{-1}$ holds $0.4\text{ kg}$ of a liquid. When $9.6\text{ kJ}$ of heat energy is supplied to the system, the temperature of the container and liquid rises by $20\text{ K}$ . What is the specific heat capacity of the liquid?

1000\text{ J kg}^{-1}\text{ K}^{-1}

1200\text{ J kg}^{-1}\text{ K}^{-1}

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

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

Show answer & explanation

Answer: $1000\text{ J kg}^{-1}\text{ K}^{-1}$

Answer

1000\text{ J kg}^{-1}\text{ K}^{-1}

The option specifying

1000\text{ J kg}^{-1}\text{ K}^{-1}

is correct because out of the

9600\text{ J}

supplied,

1600\text{ J}

(

80 \times 20

) is absorbed by the container. The remaining

8000\text{ J}

raises the temperature of

0.4\text{ kg}

of liquid by

20\text{ K}

, yielding

c = \frac{8000}{0.4 \times 20} = 1000\text{ J kg}^{-1}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the thermal energy absorbed by the container

Q_{\text{container}} = 1600\text{ J}

The heat absorbed by a body of heat capacity

C

during a temperature change

\Delta T

is given by

Q_{\text{container}} = C \Delta T = 80\text{ J K}^{-1} \times 20\text{ K} = 1600\text{ J}

.

2

Determine the net thermal energy absorbed by the liquid

Q_{\text{liquid}} = 8000\text{ J}

By energy conservation,

Q_{\text{total}} = Q_{\text{container}} + Q_{\text{liquid}}

, so

Q_{\text{liquid}} = 9600\text{ J} - 1600\text{ J} = 8000\text{ J}

.

3

Calculate the specific heat capacity of the liquid

c_{\text{liquid}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}

Using the specific heat capacity relationship

Q = m c \Delta T

,

c_{\text{liquid}} = \frac{Q_{\text{liquid}}}{m \Delta T} = \frac{8000\text{ J}}{0.4\text{ kg} \times 20\text{ K}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}

.

Key Concept

Distinction and calculation involving heat capacity of a container and specific heat capacity of a substance

Question 45Question →

The specific heat capacity of a uniform copper sphere of mass $2\text{ kg}$ is twice that of a uniform copper sphere of mass $1\text{ kg}$ at the same temperature.

True

False

Show answer & explanation

Answer: False

Answer

False. Specific heat capacity is an intensive property of a substance and is independent of the mass of the object.

The statement is false because specific heat capacity is an intensive property of matter. It measures the quantity of heat required to raise the temperature of

1\text{ kg}

of a substance by

1\text{ K}

(or

1^\circ\text{C}

). Because it is normalized per unit mass, specific heat capacity depends only on the chemical nature and state of the material, not on the total mass of the object.

Step-by-Step Solution

1

Distinguish between heat capacity and specific heat capacity.

Heat capacity (

C = mc

) is an extensive property that depends on the mass of the body, whereas specific heat capacity (

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

) is an intensive property characteristic of the material itself.

To identify whether mass affects specific heat capacity.

2

Apply the concept of intensive properties to the two copper spheres.

Since both spheres are composed of the same material (pure copper), their specific heat capacities are identical regardless of their difference in mass.

Intensive physical properties do not change with mass or sample size.

Key Concept

Intensive vs Extensive Thermal Properties

Question 46Question →

A solid sample of mass $0.25\text{ kg}$ at its melting point is completely melted by an electric heater rated at $700\text{ W}$ . If the specific latent heat of fusion of the substance is $3.36 \times 10^5\text{ J kg}^{-1}$ , what is the time taken to melt the sample completely at a constant temperature?

120\text{ s}

420\text{ s}

210\text{ s}

1200\text{ s}

Show answer & explanation

Answer: $120\text{ s}$

Answer

The time taken to completely melt the block is

120\text{ s}

.

The thermal energy required to melt a substance at its melting point depends solely on its mass and specific latent heat of fusion (

Q = m L_f

). Calculating

Q = 0.25\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 84,000\text{ J}

and dividing by heater power

700\text{ W}

yields exactly

120\text{ s}

.

Step-by-Step Solution

1

Calculate the total thermal energy required for phase change.

Q = m L_f = 0.25\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 84,000\text{ J}

During a change of state at constant temperature, thermal energy is given by

Q = m L_f

.

2

Relate energy supplied by the electrical heater to time using power formula.

Q = P \times t \implies t = \frac{Q}{P}

Power is the rate of energy transfer per unit time (

P = \frac{Q}{t}

).

3

Substitute the energy and power values to solve for time

t

.

t = \frac{84,000\text{ J}}{700\text{ W}} = 120\text{ s}

Dividing total energy required by power gives time in seconds.

Key Concept

Latent Heat of Fusion and Electrical Thermal Energy

Question 47Question →

A solid metal block absorbs $3600\text{ J}$ of thermal energy when its temperature increases from $20^\circ\text{C}$ to $60^\circ\text{C}$ . If the block has a mass of $2.0\text{ kg}$ , what is the heat capacity of the block?

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

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

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

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

Show answer & explanation

Answer: $90\text{ J K}^{-1}$

Answer

The heat capacity of the block is

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

.

The correct answer is

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

because heat capacity is defined as

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

. Dividing the absorbed heat of

3600\text{ J}

by the temperature change of

40\text{ K}

yields

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

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

)

\Delta T = 60^\circ\text{C} - 20^\circ\text{C} = 40\text{ K}

Heat capacity depends on the change in temperature, not the initial or final temperature values.

2

Apply the formula for total heat capacity (

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

)

C = \frac{3600\text{ J}}{40\text{ K}} = 90\text{ J K}^{-1}

Heat capacity

C

measures the total thermal energy required to raise the entire body's temperature by one degree Kelvin (or Celsius).

Key Concept

Heat capacity (

C

) represents the heat quantity needed to change a body's temperature by

1\text{ K}

(

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

), whereas specific heat capacity (

c

) is heat capacity per unit mass (

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

).

Estimated Time:1m 0s

Question 48Question →

An electric heater rated at $100\text{ W}$ is used to heat a metal block of mass $2.5\text{ kg}$ for $5\text{ minutes}$ . If the temperature of the block increases from $25^\circ\text{C}$ to $45^\circ\text{C}$ and no heat energy is lost to the surroundings, what is the specific heat capacity of the metal?

Show answer & explanation

Answer: 600

Answer

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

The energy transferred by the

100\text{ W}

heater over

300\text{ seconds}

is

Q = 30,000\text{ J}

. Dividing this by the product of mass (

2.5\text{ kg}

) and temperature rise (

20\text{ K}

) yields the specific heat capacity

c = 600\text{ J kg}^{-1}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate total electrical heat energy supplied to the metal block

Q = P \times t = 100\text{ W} \times (5 \times 60\text{ s}) = 30,000\text{ J}

Heat energy supplied by an electric heater is given by power multiplied by heating time in seconds.

2

Calculate the change in temperature

\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20\text{ K}

Temperature difference is calculated by subtracting initial temperature from final temperature.

3

Solve for the specific heat capacity

c = \frac{Q}{m \Delta T} = \frac{30,000\text{ J}}{2.5\text{ kg} \times 20\text{ K}} = 600\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass by one kelvin.

Key Concept

Specific Heat Capacity and Electrical Energy

Question 49Question →

A solid substance of mass $0.50\text{ kg}$ at its melting point is supplied with thermal energy by an electric heater operating at a constant rate of $150\text{ W}$ . If it takes $4.0\text{ minutes}$ for the solid to melt completely without any change in temperature, what is the specific latent heat of fusion of the substance?

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

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

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

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

Show answer & explanation

Answer: $7.2 \times 10^4\text{ J kg}^{-1}$

Answer

The specific latent heat of fusion of the substance is

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

.

During a change of phase at constant temperature, the thermal energy supplied is used exclusively to weaken intermolecular bonds without increasing kinetic energy or temperature. The heat supplied is

Q = P \times t = 150\text{ W} \times 240\text{ s} = 36\,000\text{ J}

. Using

Q = mL

, the specific latent heat

L = \frac{Q}{m} = \frac{36\,000\text{ J}}{0.50\text{ kg}} = 7.2 \times 10^4\text{ J kg}^{-1}

.

Step-by-Step Solution

1

Convert the given heating duration into SI units (seconds).

t = 4.0\text{ minutes} = 4.0 \times 60\text{ s} = 240\text{ s}

.

Standard calculation of thermal energy requires time to be in seconds when power is given in Watts.

2

Calculate the total thermal energy supplied by the heater.

Q = P \times t = 150\text{ W} \times 240\text{ s} = 36\,000\text{ J} = 3.6 \times 10^4\text{ J}

.

Electrical energy supplied is completely converted into heat energy.

3

Apply the latent heat formula to find the specific latent heat of fusion.

L = \frac{Q}{m} = \frac{36\,000\text{ J}}{0.50\text{ kg}} = 72\,000\text{ J kg}^{-1} = 7.2 \times 10^4\text{ J kg}^{-1}

.

During a change of state at constant temperature, heat absorbed is given by

Q = mL

.

Key Concept

Latent Heat of Fusion

Estimated Time:1m 30s

Question 50Question →

At a weather recording station, the air temperature is $25^\circ\text{C}$ and the relative humidity is recorded as $60\%$ . If the saturated vapour pressure of water at $25^\circ\text{C}$ is $24.0\text{ mmHg}$ , what is the actual partial vapour pressure of water present in the atmosphere?

14.4\text{ mmHg}

40.0\text{ mmHg}

9.6\text{ mmHg}

24.0\text{ mmHg}

Show answer & explanation

Answer: $14.4\text{ mmHg}$

Answer

The actual partial vapour pressure of water present in the atmosphere is

14.4\text{ mmHg}

.

Relative humidity is defined as the ratio of actual partial vapour pressure to saturated vapour pressure at the same temperature, expressed as a percentage. Therefore, the actual partial vapour pressure is

60\%

of

24.0\text{ mmHg}

, which equals

14.4\text{ mmHg}

.

Step-by-Step Solution

1

State the standard relationship defining relative humidity in terms of vapour pressures.

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

Relative humidity quantifies how close the air is to maximum moisture saturation at a given temperature.

2

Substitute the given numerical values into the relative humidity formula.

60\% = \frac{P}{24.0\text{ mmHg}} \times 100\%

Here

P

represents the unknown actual partial vapour pressure of water in the atmosphere.

3

Rearrange the equation to isolate and solve for

P

.

P = 0.60 \times 24.0\text{ mmHg} = 14.4\text{ mmHg}

Multiplying the saturated vapour pressure by

0.60

yields the actual partial vapour pressure.

Key Concept

Relative Humidity and Vapour Pressure Calculation

Question 51Question →

A glass window pane has an area of $1.50\text{ m}^2$ and a thickness of $4.0\text{ mm}$ . The inner and outer surface temperatures of the glass are maintained at $25^\circ\text{C}$ and $5^\circ\text{C}$ respectively. If the thermal conductivity of glass is $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , what is the rate of heat transfer through the glass pane by conduction?

6000\text{ W}

4000\text{ W}

87900\text{ W}

6.0\text{ W}

Show answer & explanation

Answer: $6000\text{ W}$

Answer

The rate of heat transfer through the glass pane is

6000\text{ W}

.

According to Fourier's law of thermal conduction, the rate of heat flow

\frac{Q}{t}

is given by

\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}

. Substituting

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.50\text{ m}^2

,

T_1 - T_2 = 20\text{ K}

, and

d = 0.004\text{ m}

gives

\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{0.004} = 6000\text{ W}

.

Step-by-Step Solution

1

Identify the given values and convert all quantities to SI units.

Area

A = 1.50\text{ m}^2

, thermal conductivity

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, thickness

d = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}

, and temperature difference

\Delta T = 25^\circ\text{C} - 5^\circ\text{C} = 20\text{ K}

.

Fourier's law requires all linear dimensions to be in meters and temperatures in kelvins/degrees Celsius consistently.

2

Apply Fourier's law of thermal conduction equation.

\frac{Q}{t} = \frac{k A \Delta T}{d}

The rate of heat conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to material thickness.

3

Substitute the values into the formula and solve.

\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{4.0 \times 10^{-3}} = \frac{24}{0.004} = 6000\text{ W}

Calculates the total heat energy conducted per second across the window pane.

Key Concept

Thermal Conduction Rate Equation

Estimated Time:1m 30s

Question 52Question →

A thermometer is calibrated on a custom scale, $X$ , where the ice point ( $0^\circ\text{C}$ ) is marked as $-10^\circ\text{X}$ and the steam point ( $100^\circ\text{C}$ ) is marked as $110^\circ\text{X}$ . What is the reading on this custom scale when a standard Celsius thermometer reads $35^\circ\text{C}$ ?

Show answer & explanation

Answer: 32

Answer

32 °X

Using the linear temperature interpolation formula

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{C - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}

, substituting

\text{LFP}_X = -10

,

\text{UFP}_X = 110

,

C = 35

,

\text{LFP}_C = 0

, and

\text{UFP}_C = 100

gives

\frac{X - (-10)}{110 - (-10)} = \frac{35 - 0}{100 - 0}

. Simplifying gives

\frac{X + 10}{120} = 0.35

, leading to

X + 10 = 42

, so

X = 32^\circ\text{X}

.

Step-by-Step Solution

1

Determine fundamental intervals for both temperature scales

Celsius fundamental interval =

100 - 0 = 100^\circ\text{C}

; Custom scale fundamental interval =

110 - (-10) = 120^\circ\text{X}

Linear temperature scale interpolation requires calculating the total interval between the lower fixed point (LFP) and upper fixed point (UFP).

2

Set up the ratio equation between the two thermometric scales

\frac{X - (-10)}{120} = \frac{35 - 0}{100}

The fractional position of any given temperature relative to its fixed points must be equal on all linear scales.

3

Solve the algebraic equation for

X

X + 10 = 120 \times 0.35 = 42 \implies X = 32^\circ\text{X}

Isolating

X

gives the corresponding reading on the custom temperature scale.

Key Concept

Linear Temperature Scale Conversion and Interpolation

Question 53Question →

At a constant temperature, doubling the volume of a sealed container holding a liquid and its saturated vapour will cause the saturated vapour pressure to be halved.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Saturated vapour pressure depends exclusively on temperature and is independent of the container volume.

Saturated vapour pressure is entirely independent of the volume of the space occupied by the vapour. As long as liquid remains in the container to evaporate, expanding the volume simply causes more liquid to evaporate until the vapour pressure reaches its equilibrium saturation value at that temperature.

Step-by-Step Solution

1

Analyze the physical state of a saturated vapour in contact with its liquid.

A saturated vapour exists in dynamic equilibrium with its liquid phase, meaning the rate of evaporation equals the rate of condensation at that specific temperature.

Dynamic equilibrium determines the equilibrium vapour pressure above a liquid surface.

2

Examine the effect of increasing the container volume at constant temperature.

Expanding the volume temporarily reduces vapour concentration, causing the rate of evaporation to exceed condensation until the space is re-saturated.

Phase change allows the mass of the gas phase to change, unlike in closed ideal gas systems.

3

Conclude the value of the final vapour pressure.

The pressure returns to the exact same saturated vapour pressure (SVP) value as before the expansion.

SVP is an intrinsic property dependent only on temperature and the identity of the liquid, independent of volume.

Key Concept

Independence of Saturated Vapour Pressure from Volume

Question 54Question →

A closed vessel initially contains air saturated with water vapour at $27^\circ\text{C}$ under a total pressure of $1.04 \times 10^5\text{ Pa}$ . Given that the saturated vapour pressure of water at $27^\circ\text{C}$ is $0.04 \times 10^5\text{ Pa}$ , what is the final total pressure inside the vessel if its volume is compressed to half of its initial volume while maintaining the temperature at $27^\circ\text{C}$ ?

2.04 \times 10^5\text{ Pa}

2.08 \times 10^5\text{ Pa}

2.00 \times 10^5\text{ Pa}

1.08 \times 10^5\text{ Pa}

Show answer & explanation

Answer: $2.04 \times 10^5\text{ Pa}$

Answer

The final total pressure inside the vessel is

2.04 \times 10^5\text{ Pa}

.

The total pressure is the sum of the partial pressures of dry air and saturated water vapour. Under isothermal compression to half volume, dry air follows Boyle's law and its partial pressure doubles from

1.00 \times 10^5\text{ Pa}

to

2.00 \times 10^5\text{ Pa}

. The saturated water vapour pressure remains constant at

0.04 \times 10^5\text{ Pa}

because liquid condenses out. Summing these yields

2.04 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Calculate the initial partial pressure of the dry air.

P_{\text{air, 1}} = P_{\text{total, 1}} - P_{\text{vapour, 1}} = 1.04 \times 10^5\text{ Pa} - 0.04 \times 10^5\text{ Pa} = 1.00 \times 10^5\text{ Pa}

.

According to Dalton's law of partial pressures, the total pressure of a gas mixture is the sum of the partial pressures of its individual components.

2

Determine the final partial pressure of dry air after isothermal compression.

P_{\text{air, 2}} = P_{\text{air, 1}} \times \frac{V_1}{V_2} = 1.00 \times 10^5\text{ Pa} \times 2 = 2.00 \times 10^5\text{ Pa}

.

Dry air behaves as an ideal gas and follows Boyle's law (

P_1 V_1 = P_2 V_2

) at constant temperature.

3

Determine the final partial pressure of the saturated water vapour.

P_{\text{vapour, 2}} = 0.04 \times 10^5\text{ Pa}

.

Saturated vapour pressure depends strictly on temperature. When compressed at constant temperature, excess vapour condenses into liquid, keeping the partial pressure constant at the saturated value.

4

Sum the final partial pressures to find the new total pressure.

P_{\text{total, 2}} = P_{\text{air, 2}} + P_{\text{vapour, 2}} = 2.00 \times 10^5\text{ Pa} + 0.04 \times 10^5\text{ Pa} = 2.04 \times 10^5\text{ Pa}

.

The total final pressure is the sum of the new dry air pressure and the unchanged saturated vapour pressure.

Key Concept

Saturated Vapour Pressure and Gas Law Applications

Estimated Time:2m 0s

Question 55Question →

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 56Question →

The saturated vapour pressure of water at the dew point of a mass of air is $12\text{ mmHg}$ , while the saturated vapour pressure at the actual air temperature is $24\text{ mmHg}$ . Calculate the relative humidity of the air.

Show answer & explanation

Answer: 50

Answer

The relative humidity of the air is 50%.

Relative humidity is the ratio of the saturated vapour pressure at the dew point to the saturated vapour pressure at the actual air temperature, expressed as a percentage: (12 mmHg / 24 mmHg) * 100% = 50%.

Step-by-Step Solution

1

Identify the saturated vapour pressure at the dew point and at the air temperature.

SVP at dew point = 12 mmHg; SVP at air temperature = 24 mmHg.

Relative humidity relies on the ratio of partial vapour pressure (SVP at dew point) to maximum vapour pressure at air temperature.

2

Apply the relative humidity formula.

Relative Humidity = (SVP at dew point / SVP at air temperature) * 100%

This formula defines the percentage saturation of the air.

3

Substitute the values and evaluate.

(12 / 24) * 100% = 50%

Dividing 12 by 24 gives 0.5, which equals 50% when multiplied by 100.

Key Concept

Relative Humidity Calculation

Estimated Time:45s

Question 57Question →

A constant-volume gas thermometer registers a pressure of $50\,\text{kPa}$ at the ice point ( $0^\circ\text{C}$ ) and $70\,\text{kPa}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature when the gas pressure measured by the thermometer is $62\,\text{kPa}$ ?

12^\circ\text{C}

60^\circ\text{C}

89^\circ\text{C}

310^\circ\text{C}

Show answer & explanation

Answer: $60^\circ\text{C}$

Answer

60^\circ\text{C}

The temperature

\theta

on the Celsius scale is given by the ratio of the change in thermometric property from the ice point to the total fundamental interval, scaled by 100. Substituting

P_0 = 50\,\text{kPa}

,

P_{100} = 70\,\text{kPa}

, and

P_\theta = 62\,\text{kPa}

gives

\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}

.

Step-by-Step Solution

1

Identify the given thermometric property values at the fixed points and target state

P_0 = 50\,\text{kPa}

,

P_{100} = 70\,\text{kPa}

, and

P_\theta = 62\,\text{kPa}

These represent the lower fixed point, upper fixed point, and unknown temperature reading respectively.

2

Apply the general thermometric scale conversion formula

\theta = \frac{P_\theta - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

Temperature on the Celsius scale varies linearly with the thermometric property relative to fixed points.

3

Substitute the values into the formula and solve for

\theta

\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}

Carrying out the arithmetic yields the exact temperature of

60^\circ\text{C}

.

Key Concept

Linear interpolation on temperature scales using thermometric properties

Estimated Time:1m 0s

Question 58Question →

A thermometric property $X$ of a system has a value of $15.0\,\text{units}$ at the ice point ( $0^\circ\text{C}$ ) and $75.0\,\text{units}$ at the steam point ( $100^\circ\text{C}$ ). If the thermometric property is measured as $33.0\,\text{units}$ when immersed in a chemical bath, what is the temperature of the bath on the absolute thermodynamic scale?

303\,\text{K}

317\,\text{K}

297\,\text{K}

30\,\text{K}

Show answer & explanation

Answer: $303\,\text{K}$

Answer

303\,\text{K}

The temperature of the bath on the Celsius scale is found by taking the ratio of the change in thermometric property from the ice point to the total fundamental interval:

\frac{33.0 - 15.0}{75.0 - 15.0} \times 100^\circ\text{C} = \frac{18}{60} \times 100 = 30^\circ\text{C}

. Converting this temperature to the absolute (Kelvin) scale requires adding

273\,\text{K}

, yielding

30 + 273 = 303\,\text{K}

.

Step-by-Step Solution

1

Identify the given thermometric values for the fixed points and the unknown state

X_0 = 15.0\,\text{units}

(ice point,

0^\circ\text{C}

),

X_{100} = 75.0\,\text{units}

(steam point,

100^\circ\text{C}

), and

X_T = 33.0\,\text{units}

Linear thermometric interpolation requires defining the fixed reference points and the measured property value.

2

Calculate the temperature on the Celsius scale using the linear scale formula

\theta = \frac{X_T - X_0}{X_{100} - X_0} \times 100^\circ\text{C} = \frac{33.0 - 15.0}{75.0 - 15.0} \times 100 = \frac{18.0}{60.0} \times 100 = 30^\circ\text{C}

The temperature change relative to the fundamental interval determines the position on the Celsius scale.

3

Convert the temperature from degrees Celsius to Kelvin

T = \theta + 273 = 30 + 273 = 303\,\text{K}

The absolute thermodynamic scale (Kelvin) is shifted from the Celsius scale by adding

273\,\text{K}

.

Key Concept

Linear interpolation of temperature using thermometric properties and conversion to the absolute scale

Estimated Time:2m 0s

Question 59Question →

Match each type of thermometer with its corresponding physical thermometric property.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Liquid-in-glass thermometer matches change in length or volume of a liquid column; Constant-volume gas thermometer matches change in pressure of a gas; Resistance thermometer matches change in electrical resistance; Thermocouple matches electromotive force (e.m.f.) produced across junctions.

Each thermometer is accurately matched to the physical property that undergoes a measurable change as temperature varies.

Step-by-Step Solution

1

Identify the defining physical property that varies with temperature for each instrument.

Each thermometer operates on a distinct physical property that changes predictably when heated or cooled.

Thermometric properties must be reproducible and continuously measurable across a temperature range.

2

Pair each instrument with its specific thermometric property.

Liquid-in-glass pairs with liquid column expansion/length; constant-volume gas thermometer pairs with gas pressure; resistance thermometer pairs with electrical resistance; thermocouple pairs with thermoelectric e.m.f.

These pairs represent standard physical principles used in thermometry.

Key Concept

Thermometric properties and operating principles of thermometers

Question 60Question →

Match each temperature scale reference state on the left with its correct thermodynamic definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Absolute zero matches with the state of minimum molecular kinetic energy. The ice point matches with the lower fixed point of pure melting ice at standard pressure. The steam point matches with the upper fixed point of pure boiling water steam at standard pressure. The triple point of water matches with the thermodynamic equilibrium state of ice, liquid water, and water vapour.

Each temperature scale reference point is correctly paired with its defining physical state: absolute zero represents minimum molecular kinetic energy, the ice point represents melting ice at standard pressure, the steam point represents steam from boiling water at standard pressure, and the triple point represents the three-phase equilibrium of water.

Step-by-Step Solution

1

Identify the definition of absolute zero.

Absolute zero (

0\text{ K}

) corresponds to the state of minimum internal molecular kinetic energy.

At

0\text{ K}

, thermal motion of particles theoretically ceases.

2

Identify the definition of the ice point.

The ice point (

0^\circ\text{C}

) corresponds to pure melting ice at standard atmospheric pressure.

It serves as the standard lower fixed point on the Celsius temperature scale.

3

Identify the definition of the steam point.

The steam point (

100^\circ\text{C}

) corresponds to steam from pure boiling water at standard atmospheric pressure.

It serves as the standard upper fixed point on the Celsius temperature scale.

4

Identify the definition of the triple point of water.

The triple point (

273.16\text{ K}

) is the unique thermodynamic state where ice, liquid water, and steam coexist in equilibrium.

It is used as a single fundamental reference point on the Kelvin thermodynamic scale.

Key Concept

Temperature Scale Fixed Points and Reference States

Estimated Time:1m 30s

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