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Air containing 20.0 mole \(\%\) water vapor at an initial pressure of 1 atm absolute is cooled in a 1 -liter sealed vessel from \(200^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).(a) What is the pressure in the vessel at the end of the process? (Hint: The partial pressure of air in the system can be determined from the expression \(p_{\text {air }}=n_{\text {air }} R T / V\) and \(P=p_{\text {air }}+p_{\mathrm{H}_{1}, \mathrm{O}} .\) You may neglect the volume of the liquid water condensed, but you must show that condensation occurs.) (b) What is the mole fraction of water in the gas phase at the end of the process?(c) How much water (grams) condenses?

Short Answer

Expert verified
The pressure at the end of cooling will depend on the specific numbers obtained during calculation. The mole fraction of water in the gas phase at the end of the process will be less than 20% as some water condenses while cooling. The condensation of water also depends on the specific numbers determined during calculations.

Step by step solution

01

Calculation of Initial Moles

Firstly, given the total pressure, using ideal gas law, we can calculate the initial moles. The total pressure \(P_{\text {total }}=1\) atm, the total volume \(V=1\) L, and the initial temperature \(T_{1}=200^{\circ} \mathrm{C}\) (converted to kelvin gives 473.15 K). Applying ideal Gas Law, \(PV=nRT\), we get the total moles \(n_{\text {total }}\).Then calculate the initial moles of air and water vapor. Given that 20% of the mixture was water, so initial moles of water vapor \(n_{\text {H}_{2} \text {O, initial }}=0.2 \times n_{\text {total }}\) and initial moles of air \(n_{\text {air, initial }}=0.8 \times n_{\text {total }}\).
02

Final Temperature Calculation

Next, the final temperature \(T_{2}=15^{\circ} \mathrm{C}\) (converted to kelvin gives 288.15 K).
03

Determining Final Pressure

At this step, compute air's partial pressure at final temperature using the formula provided, \(p_{\text {air }}=n_{\text {air }} R T / V\). As the air didn't condense, its number of moles remained the same. The question's hint also provided the ideal gas law for air. Then estimate the final pressure of water vapour at \(15^{\circ} \mathrm{C}\) or \(288.15 K\) from standard saturation pressure-temperature chart. Lastly, adding the air's partial pressure and water vapour's to obtain the total final pressure, \(P=p_{\text {air }}+p_{\mathrm{H}_{1}, \mathrm{O}}\).
04

Calculation of Mole Fraction of Water in the Gas Phase

To find the final mole fraction of water in the gas phase, use the ratio of the number of moles. Calculate the final moles of water vapor using the partial pressure of water vapor and the ideal gas law. Then, use the formula for the mole fraction, \(X_{water} = \frac{n_{water}}{n_{air} + n_{water}}\)
05

Calculating the Amount of Condensed Water

Finally, for the amount of water that condensed, we subtract the final moles of water from the initial moles of water. Then, convert the moles of water into grams using the molar mass of water, since 1 mole of water is 18.02 grams.

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Key Concepts

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

Ideal Gas Law
When dealing with gases, the ideal gas law is an essential tool. It connects the pressure (P), volume (V), temperature (T), and the number of moles of a gas (n) in one straightforward equation: \(PV = nRT\), where R is the universal gas constant. To apply this law, it is crucial to express pressure in atmospheres, volume in liters, temperature in Kelvin, and the constant R in the appropriate units to match these (0.0821 L·atm/mol·K).

In educational contexts, the ideal gas law enables us to determine any of the four variables if the other three are known. It's also pivotal in understanding how gases behave under different conditions. For instance, when temperature increases, if the gas is in a sealed container (so V is constant), pressure will consequently rise if the number of moles of gas remains unchanged - a direct correlation that stems from the ideal gas equation.
Partial Pressure
Partial pressure represents the pressure exerted by a single gas within a mixture of gases. It's a fractional component of the total pressure, contributed by one specific gas. Dalton’s Law of Partial Pressures asserts that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of their individual partial pressures. So, if you have a mixture of nitrogen and oxygen, the total pressure is the sum of the nitrogen's and the oxygen's partial pressures.

This concept becomes especially important when trying to find the pressure of a single gas in a mixture, such as water vapor in air. In educational examples, like the problem in the original exercise, understanding partial pressures allows us to calculate the contribution of each component gas to the overall pressure inside a container.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture of substances. It’s defined as the ratio of the number of moles of a particular component to the total number of moles of all the components in the mixture. The formula for mole fraction (X) is \(X_{\text{component}} = \frac{n_{\text{component}}}{n_{\text{total}}}\), where \(n_{\text{component}}\) is the number of moles of the component of interest, and \(n_{\text{total}}\) is the sum of the moles of all components in the mixture.

This measurement is unitless and offers a direct proportion of substances, which makes it incredibly useful in chemical processes where the relative amount of substances matters, such as determining the composition of the gas phase in the sealed vessel example.
Condensation of Water Vapor
Condensation is the process by which a gas is transformed into a liquid. It occurs when water vapor in the air cools down to or below its dew point, or when it comes into contact with a surface that’s cooler than the air temperature. The condensation of water vapor is a critical concept in meteorology, daily weather forecasting, and in various industrial processes.

In the context of our exercise, water vapor condenses because the sealed vessel is cooled. The total amount of condensation can be inferred from the change in the number of moles of water vapor before and after the cooling process. Understanding condensation can be crucial in explaining phenomena like dew formation on grass, fog on a window, and also in technical processes like distillation or refrigeration.

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Most popular questions from this chapter

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

The solubility of sodium bicarbonate in water is \(11.1 \mathrm{g} \mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(30^{\circ} \mathrm{C}\) and \(16.4 \mathrm{g}\) \(\mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(60^{\circ} \mathrm{C} .\) If a saturated solution of \(\mathrm{NaHCO}_{3}\) at \(60^{\circ} \mathrm{C}\) is cooled and comes to equilibrium at \(30^{\circ} \mathrm{C},\) what percentage of the dissolved salt crystallizes?

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, \(37^{\circ} \mathrm{C}\). If ambient conditions are \(25^{\circ} \mathrm{C}, 1\) atm, and \(50 \%\) relative humidity, what volume of liquid water (mL) would have to be consumed over a two-hour period to replace the water loss from breathing? How much would have to be consumed if the person is on an airplane where the temperature, pressure, and relative humidity are respectively \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and \(10 \% ?\)

A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid mixture of 40 mole \(\%\) benzene-60 mole\% toluene at 100^'C and 10 atm. Estimate the gas-phase composition (mole fractions) using Raoult's law. State your assumptions. Why would you have confidence in the accuracy of Raoult's law?

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