Vapour Pressure, Boiling, Evaporation, and Relative Humidity Practice Questions — JAMB UTME

Question 1Question →

On a warm afternoon, the air temperature in a physics laboratory is $30^\circ\text{C}$ , where the saturated vapour pressure of water is $32.0\text{ mmHg}$ . When the air is cooled, condensation just begins to form on a metal vessel at $20^\circ\text{C}$ . Given that the saturated vapour pressure of water at $20^\circ\text{C}$ is $17.6\text{ mmHg}$ , what is the relative humidity of the air in percentage?

Show answer & explanation

Answer: 55

Answer

The relative humidity of the air is

55\%

.

The dew point is the temperature at which condensation begins, indicating that the actual water vapour pressure present in the air equals the saturated vapour pressure at

20^\circ\text{C}

, which is

17.6\text{ mmHg}

. Dividing this actual vapour pressure by the saturated vapour pressure at the ambient air temperature of

30^\circ\text{C}

(

32.0\text{ mmHg}

) and multiplying by

100\%

yields

\frac{17.6}{32.0} \times 100\% = 55\%

.

Step-by-Step Solution

1

Determine the actual vapour pressure in the air

Actual vapour pressure =

17.6\text{ mmHg}

Condensation starts at the dew point (

20^\circ\text{C}

), meaning the actual vapour pressure in the air equals the saturated vapour pressure at the dew point.

2

Determine the saturated vapour pressure at the air temperature

Saturated vapour pressure at

30^\circ\text{C}

=

32.0\text{ mmHg}

This is the maximum vapour pressure the air can exert at its current ambient temperature.

3

Compute the relative humidity percentage

\text{Relative Humidity} = \frac{17.6}{32.0} \times 100\% = 55\%

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

Key Concept

Calculation of relative humidity from saturated vapour pressure at dew point and air temperature

Estimated Time:1m 30s

Question 2Question →

A liquid will boil when its saturated vapour pressure becomes equal to the prevailing external atmospheric pressure.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure.

The statement accurately expresses the fundamental thermodynamic condition for boiling: the temperature of the liquid must reach a point where its saturated vapour pressure equals the surrounding atmospheric pressure.

Step-by-Step Solution

1

Recall the definition of boiling point in thermal physics.

Boiling is the rapid conversion of liquid into gas occurring throughout the liquid body.

To determine the exact physical condition required for boiling to take place.

2

Relate saturated vapour pressure (SVP) to atmospheric pressure.

Bubbles of vapour can form within the liquid only when the pressure inside the bubbles (SVP) is equal to or greater than the pressure pushing down from the outside atmosphere.

If SVP is lower than atmospheric pressure, any vapour bubble attempting to form inside the liquid will immediately collapse.

Key Concept

Condition for Boiling of Liquids

Question 3Question →

When air containing unsaturated water vapour is cooled at constant atmospheric pressure without adding or removing moisture, the saturated vapour pressure of water decreases while the actual partial vapour pressure of water remains constant until the dew point is reached.

True

False

Show answer & explanation

Answer: True

Answer

True. Saturated vapour pressure decreases as temperature falls, while actual partial vapour pressure remains constant under constant total pressure until condensation starts at the dew point.

The statement is correct because saturated vapour pressure is a temperature-dependent property that decreases as air cools. So long as total pressure remains unchanged and no water vapour is added or removed, the actual partial vapour pressure of the water vapour stays constant until the dew point is reached, at which point the air becomes saturated (

100\%

relative humidity) and condensation begins.

Step-by-Step Solution

1

Analyze the temperature dependence of saturated vapour pressure (SVP).

SVP decreases as temperature drops.

SVP is determined by the kinetic energy of water molecules escaping into vapour at dynamic equilibrium, which decreases with decreasing temperature.

2

Determine the behavior of the actual partial vapour pressure of water during cooling at constant total pressure.

Actual partial vapour pressure remains constant prior to condensation.

Dalton's law dictates that partial pressure depends on the mole fraction of water vapour and total pressure; since no water vapour is added or removed, actual partial vapour pressure does not change.

3

Evaluate the condition for reaching the dew point.

At the dew point, SVP drops to equal the constant actual partial vapour pressure, achieving 100% relative humidity.

Relative humidity is defined as

\text{R.H.} = \frac{P_{\text{actual}}}{P_{\text{SVP}}} \times 100\%

; as

P_{\text{SVP}}

decreases towards

P_{\text{actual}}

,

\text{R.H.}

increases to

100\%

.

Key Concept

Saturated vs. actual vapour pressure, temperature dependence of SVP, and dew point

Question 4Question →

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 5Question →

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 6Question →

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 7Question →

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 8Question →

The boiling point of a liquid decreases when the external atmospheric pressure acting on its surface is reduced.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Boiling occurs when the saturated vapour pressure of a liquid equals the external atmospheric pressure. Decreasing the external atmospheric pressure reduces the saturated vapour pressure threshold required for boiling, allowing the liquid to boil at a lower temperature.

Step-by-Step Solution

1

Identify the condition required for boiling to occur.

Boiling takes place when the saturated vapour pressure of the liquid equals the external atmospheric pressure.

This is the fundamental physical definition of boiling.

2

Analyze the effect of reducing external atmospheric pressure.

Lower external pressure means the required saturated vapour pressure is reached at a lower temperature.

Saturated vapour pressure increases with temperature, so a lower target pressure corresponds to a lower boiling temperature.

Key Concept

Dependence of Boiling Point on External Atmospheric Pressure

Question 9Question →

If a sample of unsaturated air with a relative humidity of $40\%$ at $30^\circ\text{C}$ is compressed isothermally to one-third of its original volume, the final relative humidity of the air will be $100\%$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The relative humidity cannot exceed 100% because water vapour condenses into liquid once its partial pressure reaches the saturated vapour pressure at that temperature.

The statement is true because relative humidity is the ratio of actual partial vapour pressure to saturated vapour pressure at a specific temperature. Although isothermal compression to one-third volume triples the partial pressure of water vapour (giving a theoretical 120%), partial vapour pressure is bounded by the saturated vapour pressure. When the relative humidity reaches 100%, condensation occurs, keeping the relative humidity at exactly 100%.

Step-by-Step Solution

1

Apply Boyle's law to calculate the theoretical partial vapour pressure after isothermal compression.

Since volume is reduced to

\frac{1}{3}V_1

at constant temperature, the partial pressure of water vapour would increase by a factor of 3 (

p_2 = 3p_1

).

For an unsaturated vapour at constant temperature, partial pressure is inversely proportional to volume.

2

Determine the theoretical relative humidity from the pressure ratio.

\text{Theoretical R.H.} = 3 \times 40\% = 120\%.

Relative humidity is defined as

\text{R.H.} = \frac{p}{p_s} \times 100\%

, where

p

is the partial pressure and

p_s

is the saturated vapour pressure.

3

Apply the physical limit imposed by saturated vapour pressure.

The actual relative humidity cannot exceed

100\%

; excess vapour condenses.

Saturated vapour pressure

p_s

represents the maximum possible partial pressure of water vapour in air at a given temperature.

Key Concept

Saturated Vapour Pressure Limit and Relative Humidity under Isothermal Compression

Question 10Question →

A mass of air at $30^\circ\text{C}$ has a relative humidity of $50\%$ . The saturated vapour pressure of water is $30\text{ mmHg}$ at $30^\circ\text{C}$ and $9\text{ mmHg}$ at $10^\circ\text{C}$ . If the temperature of the air is lowered to $10^\circ\text{C}$ , what percentage of the initial water vapour condenses out?

$40\%

$60\%

$70\%

$30\%

Show answer & explanation

Answer: $40\%

Answer

The percentage of initial water vapour that condenses out when cooled to

10^\circ\text{C}

is

40\%

.

The correct answer is

40\%

. Initially, the air contains water vapour exerting a partial pressure of

0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

. Upon cooling to

10^\circ\text{C}

, the air becomes saturated at

9\text{ mmHg}

, causing

15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

worth of vapour to condense into liquid. The condensed amount as a fraction of the initial vapour is

6 / 15 = 0.40

, or

40\%

.

Step-by-Step Solution

1

Calculate the initial partial vapour pressure of water at

30^\circ\text{C}

.

\text{Partial Vapour Pressure} = \text{Relative Humidity} \times \text{SVP at } 30^\circ\text{C} = 0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

.

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the amount of vapour pressure that must condense when cooled to

10^\circ\text{C}

.

\text{Vapour pressure condensed} = 15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

.

At

10^\circ\text{C}

, the air can hold at most its saturated vapour pressure of

9\text{ mmHg}

, so any excess vapour above

9\text{ mmHg}

condenses into liquid water.

3

Calculate the percentage of the initial water vapour that condenses out.

\text{Percentage condensed} = \left(\frac{6\text{ mmHg}}{15\text{ mmHg}}\right) \times 100\% = 40\%

.

The question asks for the fraction of the initial vapour originally present that leaves the gaseous state.

Key Concept

Relative Humidity and Dew Point Condensation

Estimated Time:1m 30s

Question 11Question →

A closed rigid container holds a mixture of dry air and water vapour at a temperature of $27^\circ\text{C}$ under a total pressure of $740\text{ mmHg}$ . The relative humidity of the air inside the container is $80\%$ , and the saturated vapour pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If the container is heated at constant volume to $127^\circ\text{C}$ , what is the partial pressure of the dry air in $\text{mmHg}$ at this higher temperature?

Show answer & explanation

Answer: 960

Answer

The partial pressure of dry air inside the container at

127^\circ\text{C}

is

960\text{ mmHg}

.

First, the partial pressure of water vapour at

27^\circ\text{C}

is determined by multiplying relative humidity (

80\%

) by the saturated vapour pressure (

25\text{ mmHg}

), yielding

20\text{ mmHg}

. Next, subtracting this vapour pressure from the total pressure of

740\text{ mmHg}

gives the partial pressure of dry air alone as

720\text{ mmHg}

at

27^\circ\text{C}

(

300\text{ K}

). Finally, applying the Pressure Law (

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

) for the dry air between

300\text{ K}

and

400\text{ K}

(

127^\circ\text{C}

) yields

P_2 = 720 \times \frac{400}{300} = 960\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the partial pressure of water vapour at

27^\circ\text{C}

P_{\text{vapour}, 1} = 0.80 \times 25\text{ mmHg} = 20\text{ mmHg}

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the initial partial pressure of the dry air at

27^\circ\text{C}

using Dalton's Law

P_{\text{dry air}, 1} = 740\text{ mmHg} - 20\text{ mmHg} = 720\text{ mmHg}

Total pressure of a gas mixture is the sum of the partial pressures of its individual components.

3

Convert temperatures to Kelvin and apply Gay-Lussac's Pressure Law for dry air at constant volume

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

,

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

.

P_{\text{dry air}, 2} = 720\text{ mmHg} \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 960\text{ mmHg}

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

Key Concept

Dalton's Law of Partial Pressures and Gay-Lussac's Pressure Law applied to gas-vapour mixtures

Question 12Question →

A sealed room with a volume of $50\text{ m}^3$ at a temperature of $20^\circ\text{C}$ contains $0.40\text{ kg}$ of water vapour. If the mass of water vapour required to saturate $1\text{ m}^3$ of air at $20^\circ\text{C}$ is $0.016\text{ kg}$ , what is the relative humidity of the air in the room?

25\%

40\%

50\%

80\%

Show answer & explanation

Answer: $50\%$

Answer

50\%

Relative humidity is defined as the ratio of the actual mass of water vapour present in a given volume of air to the mass of water vapour required to saturate the same volume at the same temperature. For

50\text{ m}^3

of air, the mass needed for saturation is

50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

. Dividing the actual mass (

0.40\text{ kg}

) by

0.80\text{ kg}

and multiplying by

100\%

gives

50\%

.

Step-by-Step Solution

1

Calculate the total mass of water vapour required to saturate the entire room volume.

\text{Total saturation mass} = 50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

The saturation density gives the maximum water vapour

1\text{ m}^3

can hold, so it must be scaled by the room volume.

2

Apply the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Actual mass of water vapour}}{\text{Saturation mass of water vapour}}\right) \times 100\%

Relative humidity measures the degree of saturation of an air sample at a specific temperature.

3

Substitute the known values to find the relative humidity.

\text{Relative Humidity} = \left(\frac{0.40\text{ kg}}{0.80\text{ kg}}\right) \times 100\% = 50\%

Simplifying the fraction

\frac{0.40}{0.80} = 0.5

, which equals

50\%

.

Key Concept

Relative Humidity

Estimated Time:1m 15s

Question 13Question →

If a fixed mass of unsaturated air in a closed container is cooled at constant total pressure, its relative humidity increases until it reaches the dew point; upon cooling further below the dew point, condensation occurs such that the relative humidity remains at 100% while the saturated vapour pressure continues to decrease.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Cooling unsaturated air reduces its maximum moisture-holding capacity (SVP). Before saturation, actual vapour pressure is fixed, so relative humidity rises to 100% at the dew point. Continuous cooling past this point forces condensation, decreasing the actual vapour pressure alongside the SVP to keep the air continuously saturated at 100% relative humidity.

Step-by-Step Solution

1

Analyze the relationship between cooling and relative humidity for unsaturated air.

As temperature decreases, the saturated vapour pressure (SVP) decreases while the actual partial vapour pressure remains constant, causing relative humidity (actual VP / SVP × 100%) to increase.

Relative humidity is inversely proportional to the saturated vapour pressure at a fixed moisture content.

2

Identify the state reached when relative humidity reaches 100%.

The air reaches saturation at the dew point temperature where actual vapour pressure equals SVP.

By definition, the dew point is the temperature at which water vapour in air begins to condense.

3

Determine the thermodynamic behavior during continuous cooling below the dew point.

Excess vapour condenses into liquid water, decreasing actual vapour pressure to continuously match the lower SVP at each reduced temperature.

Air cannot maintain an unsaturated or supersaturated equilibrium state in the presence of condensate, keeping relative humidity locked at 100%.

Key Concept

Vapour saturation, dew point determination, and condensation mechanics during air cooling.

Estimated Time:1m 30s

Question 14Question →

On a warm day, the saturated vapour pressure of water at an ambient temperature of $25^\circ\text{C}$ is $24\text{ mmHg}$ . If the actual partial pressure of water vapour in the air at this temperature is $18\text{ mmHg}$ , what is the relative humidity of the atmosphere?

75\%

133\%

25\%

43\%

Show answer & explanation

Answer: $75\%$

Answer

The relative humidity of the atmosphere is

75\%

.

The relative humidity is defined as the ratio of the partial vapour pressure of water present in the air to the saturated vapour pressure at the ambient temperature, expressed as a percentage. Dividing

18\text{ mmHg}

by

24\text{ mmHg}

yields

0.75

, which equals

75\%

.

Step-by-Step Solution

1

Identify the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Partial Vapour Pressure}}{\text{Saturated Vapour Pressure at same temp}}\right) \times 100\%

Relative humidity measures how close atmospheric air is to saturation at a given temperature.

2

Substitute the given values into the formula.

\text{Relative Humidity} = \left(\frac{18\text{ mmHg}}{24\text{ mmHg}}\right) \times 100\%

The partial pressure is

18\text{ mmHg}

and the saturated vapour pressure is

24\text{ mmHg}

.

3

Calculate the final percentage.

\text{Relative Humidity} = 0.75 \times 100\% = 75\%

Simplifying

\frac{18}{24}

gives

\frac{3}{4}

, which corresponds to

0.75

or

75\%

.

Key Concept

Relative humidity is the ratio of the actual mass (or partial pressure) of water vapour in a given volume of air to the maximum mass (or saturated vapour pressure) of water vapour that the same volume of air can hold at that same temperature, expressed as a percentage.

Question 15Question →

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 16Question →

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 17Question →

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

Vapour Pressure, Boiling, Evaporation, and Relative Humidity

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