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A gaseous mixture inside a rigid steel vessel contains \(75 \% \mathrm{CO}_{2}(g)\) and \(25 \% \mathrm{H}_{2} \mathrm{O}(g)\) by volume at \(175^{\circ} \mathrm{C}\) and \(225 \mathrm{kPa}\). The mixture is then cooled to \(0^{\circ} \mathrm{C}\), thereby condensing the water vapor. What is the pressure of the \(\mathrm{CO}_{2}(g)\) at \(0^{\circ} \mathrm{C} ?\) (Assume that there is no water vapor present after condensation.)

Short Answer

Expert verified
The pressure of CO₂ at 0°C is 102.65 kPa.

Step by step solution

01

Calculate the Initial Partial Pressure of COâ‚‚

First, determine the partial pressure of COâ‚‚ at the initial temperature and pressure. Since the mixture is 75% COâ‚‚ by volume, use the formula for partial pressure:\[ P_{ ext{CO}_2} = X_{ ext{CO}_2} \times P_{ ext{total}} \]where \( X_{ ext{CO}_2} = 0.75 \) and \( P_{ ext{total}} = 225 \ \text{kPa} \).\[ P_{ ext{CO}_2} = 0.75 \times 225 = 168.75 \ \text{kPa} \]
02

Identify the Condition Changes

Recognize that after cooling to \(0^{\circ} \text{C} \), the water vapor condenses completely, leaving only COâ‚‚ in the gas phase. Observe that, since the vessel is rigid (fixed volume), the total number of moles of COâ‚‚ remains constant, but its pressure will change since it's the only gas remaining.
03

Apply the Ideal Gas Law for COâ‚‚

Apply the ideal gas law to find the new pressure of COâ‚‚ at \(0^{\circ} \text{C}\). Since volume and the number of moles are constant:\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]Use \(P_1 = 168.75 \ \text{kPa}\), \(T_1 = 175^{\circ} \text{C} = 448 \text{K} \), and \(T_2 = 0^{\circ} \text{C} = 273 \text{K} \).Rearranging the equation, we get:\[ P_2 = P_1 \times \frac{T_2}{T_1} \]Substitute the values:\[ P_2 = 168.75 \times \frac{273}{448} = 102.65 \ \text{kPa} \]
04

Finalize the Pressure of CO₂ at 0°C

After the calculations, we determine the pressure of COâ‚‚ at \(0^{\circ} \text{C}\) is \(102.65 \ \text{kPa}\). This accounts for the fact that all water vapor has condensed, leaving the pressure due only to COâ‚‚.

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Key Concepts

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

Partial Pressure
When dealing with a gas mixture, each gas component exerts its own pressure, which is known as its partial pressure. The partial pressure is proportional to its mole fraction in the mixture. This relationship is crucial for understanding gas behavior in mixtures.
In the original exercise, we have a mixture containing 75% COâ‚‚ and 25% Hâ‚‚O by volume. To find the partial pressure of COâ‚‚, we use the formula:
  • \( P_{\text{CO}_2} = X_{\text{CO}_2} \times P_{\text{total}} \)
Here, \( X_{\text{CO}_2} = 0.75 \) is the mole fraction, and \( P_{\text{total}} = 225 \) kPa is the total pressure. Hence, the initial partial pressure of COâ‚‚ is 168.75 kPa.
Understanding partial pressure helps us segregate how much each gas in a mixture contributes to the total pressure.
Gas Mixture
A gas mixture consists of multiple gases that are mixed together but not chemically bonded. Each gas within a mixture behaves independently.
In our example, we have CO₂ and H₂O gases. These gases coexist under an initial condition of a rigid container with a total pressure of 225 kPa and a temperature of 175°C. The properties of the mixture depend on the individual proprietaries of the gases involved.
  • Each gas contributes to the total pressure.
  • The contribution is dependent on its fractional volume and partial pressure.
Analyzing gas mixtures helps in situations where gas separation or collection is required, such as environmental gas analysis and chemical reactions.
Condensation
Condensation is the process where a substance in gas form transforms into a liquid. This occurs when the gas is either cooled or compressed to reach its dew point.
In the problem provided, when the gas mixture was cooled to 0°C, the water vapor condenses. At this point, liquid water forms, leaving CO₂ as the only gas present in the vessel. This process reduces the total pressure of the system since the partial pressure contributed by the water vapor is now zero.
  • Condensation removes gas volume, affecting pressures in mixtures.
  • It helps transition between gas phases in thermodynamic cycles.
Understanding condensation is essential for studying climatic processes and designing systems like refrigerators and dehumidifiers.
Temperature Change
Temperature change impacts gas behavior significantly, altering volume and pressure conditions according to the Ideal Gas Law. Lowering the temperature of a gas usually reduces its kinetic energy, causing a decrease in pressure if the volume is constant.
In the scenario given, the temperature of the gas mixture is reduced from 175°C (448 K) to 0°C (273 K). Assuming volume and amount of CO₂ remain unchanged, the pressure change can be calculated using:
  • \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Substituting known values leads to a new COâ‚‚ pressure of 102.65 kPa. This principle helps in understanding gas expansion and contraction, crucial for engines, weather predictions, and pressure vessels.

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Most popular questions from this chapter

Convert the following temperatures to the Kelvin scale: (a) \(-183^{\circ} \mathrm{C}\) (the melting point of oxygen) (b) \(6000^{\circ} \mathrm{C}\) (temperature at the surface of the sun) (c) \(-269^{\circ} \mathrm{C}\) (the boiling point of helium) (d) \(800^{\circ} \mathrm{C}\) (the melting point of sodium chloride)

The Dumas method was one of the first reliable techniques for determining the molar mass of a volatile liquid. In this method a liquid is boiled inside an open glass bulb so that its vapor completely fills the bulb, displacing any air present. The vapor inside the bulb is then cooled and the condensate weighed. From this mass, the boiling point of the liquid, and the atmospheric pressure, one can determine the molar mass of the liquid. A chemist using this technique finds that a pure volatile liquid with a boiling point of \(80.1^{\circ} \mathrm{C}\) fills a \(500.0-\mathrm{cm}^{3}\) bulb at an atmospheric pressure of \(755.3\) Torr. If the weighed condensate has a mass of \(1.335\) grams, what is the molecular mass of the liquid? If the empirical formula of the unknown liquid is \(\mathrm{CH}\), what is the molecular formula?

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Suppose we wish to inflate a weather balloon with helium. The balloon should have a volume of \(100 \mathrm{~m}^{3}\) when inflated to a pressure of \(0.10\) bar. If we use \(50.0\) -liter cylinders of compressed helium gas at a pressure of 100 bar, how many cylinders do we need? Assume that the temperature remains constant.

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