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The vapor pressure of water at \(10^{\circ} \mathrm{C}\) is \(1300 \mathrm{Pa}\). (a) What per centage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5}\) Pa. (b) What percentage of the total air pressure at \(10^{\circ} \mathrm{C}\) is due to water vapor when the relative humidity is \(100 \% ?\) (c) The vapor pressure of water at \(35^{\circ} \mathrm{C}\) is \(5500 \mathrm{Pa}\). What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} \mathrm{C}\) when the relative humidity was \(100 \% ?\)

Short Answer

Expert verified
1.283% (a), 1.283% (b), 23.636% (c)

Step by step solution

01

Calculate the percentage of atmospheric pressure

The vapor pressure of water at \(10^\circ C\) is given as \(1300 \text{ Pa}\). Atmospheric pressure is \(1.013 \times 10^5 \text{ Pa}\). To find the percentage of atmospheric pressure that the water vapor pressure constitutes, use the formula:\[ \text{Percentage} = \left( \frac{1300}{1.013 \times 10^5} \right) \times 100 \% \]Plugging in the values, we get:\[ \text{Percentage} = \left( \frac{1300}{101300} \right) \times 100 \approx 1.283 \% \]
02

Calculate air pressure percentage due to water vapor

When the relative humidity is \(100\%\), the vapor pressure is equal to the partial pressure of the water in the air. Therefore, at \(10^\circ C\), with a total possible vapor pressure of \(1300 \text{ Pa}\), water vapor accounts for the same percentage of the total atmospheric pressure we calculated in Step 1, which is \(1.283\%\) of the atmospheric pressure.
03

Determine the relative humidity at 35°C

At \(35^\circ C\), the vapor pressure can be \(5500 \text{ Pa}\). At \(10^\circ C\), the vapor pressure was \(1300 \text{ Pa}\) with \(100\%\) humidity, meaning the partial pressure of water remains \(1300 \text{ Pa}\) at \(35^\circ C\) if it has not changed. The relative humidity at \(35^\circ C\) can be calculated using:\[ \text{Relative Humidity} = \left( \frac{1300}{5500} \right) \times 100\% \]This gives:\[ \text{Relative Humidity} = \left( \frac{1300}{5500} \right) \times 100 \approx 23.636\% \]
04

Summarize the results

(a) The vapor pressure of water at \(10^\circ C\) is approximately \(1.283\%\) of the atmospheric pressure. (b) With \(100\%\) relative humidity, water vapor contributes \(1.283\%\) to the total air pressure at \(10^\circ C\). (c) At \(35^\circ C\), the relative humidity is approximately \(23.636\%\) if the partial pressure of water in the air remains at \(1300 \text{ Pa}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relative Humidity
Relative humidity is a measure of how much moisture the air currently holds compared to the maximum amount it could hold at a particular temperature. It is expressed as a percentage.
To understand it better, think of a sponge: a completely soaked sponge represents 100% relative humidity, whereas a damp sponge represents less. Relative humidity can be calculated using the formula:\[\text{Relative Humidity} = \left( \frac{\text{Actual Vapor Pressure}}{\text{Saturation Vapor Pressure}} \right) \times 100\%\]This formula shows the relationship between the partial pressure of the water vapor currently in the air (or actual vapor pressure) and the vapor pressure at saturation (maximum, at a specific temperature).
  • If the relative humidity is 100%, it means the air is fully saturated with water vapor and can't hold more without some condensing, like dew or rain.
  • A lower percentage means the air is drier.
In the problem above, when the temperature changes but the partial pressure of water remains the same, the relative humidity changes based on its current capacity to hold moisture.
Atmospheric Pressure
Atmospheric pressure (also known as air pressure) is the force exerted by the weight of the air above a surface. It is an important concept in understanding how weather and climate work.
The atmospheric pressure is measured in units like pascal (Pa) or atmospheres (atm).It can be different based on location and elevation:
  • At sea level, the standard atmospheric pressure is about \(1.013 \times 10^5\) Pa (or 1013 hPa or 1 atm).
  • It decreases with higher altitude because there is less air above.
In the context of our exercise, knowing the atmospheric pressure helps in understanding the proportion that water vapor pressure constitutes at different temperatures. Remember, water vapor contributes to the total air pressure, and when relative humidity reaches 100%, the vapor pressure equals the partial pressure of the water vapor.
Partial Pressure
Partial pressure is the pressure contributed by a single type of gas in a mixture of gases. When we talk about air, it's made up of several gases like nitrogen, oxygen, and water vapor among others.
In atmospheric science, partial pressures are pivotal in understanding the behavior and proportionate contribution of each gas to the total atmospheric pressure.To calculate the partial pressure of a gas in a mixture, we use the formula:\[P_{i} = \frac{n_{i}}{n_{\text{total}}} \times P_{\text{total}}\]where \(P_{i}\) is the partial pressure of the gas, \(n_{i}\) is the number of moles of the gas, \(n_{\text{total}}\) is the total number of moles, and \(P_{\text{total}}\) is the total pressure of the gas mixture. For example:
  • The partial pressure of water vapor at \(10^\circ C\) in our problem was \(1300 \text{ Pa}\), forming a small part of the total atmospheric pressure.
  • In cases of 100% relative humidity, the actual vapor pressure equals the saturation vapor pressure, making partial pressure calculations straightforward in such scenarios.
Understanding partial pressure allows us to see how each component of the air affects overall pressure, crucial for fields like meteorology and climate science.

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Most popular questions from this chapter

An ice chest at a beach party contains 12 cans of soda at \(5.0^{\circ} \mathrm{C}\). Each can of soda has a mass of \(0.35 \mathrm{kg}\) and a specific heat capacity of \(3800 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) Someone adds a \(6.5-\mathrm{kg}\) watermelon at \(27^{\circ} \mathrm{C}\) to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature \(T\) of the soda and watermelon.

Two bars of identical mass are at \(25^{\circ} \mathrm{C} .\) One is made from glass and the other from another substance. The specific heat capacity of glass is \(840 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88{ }^{\circ} \mathrm{C},\) while the other bar reaches \(250.0^{\circ} \mathrm{C} .\) What is the specific heat capacity of the other substance?

A constant-volume gas thermometer (see Figures 12.3 and 12.4\()\) has a pressure of \(5.00 \times 10^{3} \mathrm{Pa}\) when the gas temperature is \(0.00^{\circ} \mathrm{C}\). What is the temperature (in \({ }^{\circ} \mathrm{C}\) ) when the pressure is \(2.00 \times 10^{3} \mathrm{Pa} ?\)

An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can's initial volume at \(5^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-4} \mathrm{m}^{3} .\) The coefficient of volume expansion for aluminum is \(69 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) When the can and the liquid are heated to \(78^{\circ} \mathrm{C}, 3.6 \times 10^{-6} \mathrm{m}^{3}\) of liquid spills over. What is the coefficient of volume expansion of the liquid?

A 0.35-kg coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C} .\) To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

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