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A \(1.25-\mathrm{g}\) sample of \(\mathrm{CO}_{2}\) is contained in a \(750 .\) -mL flask at \(22.5^{\circ} \mathrm{C} .\) What is the pressure of the gas?

Short Answer

Expert verified
The pressure is approximately 0.921 atm.

Step by step solution

01

Understanding the Problem

You are given the mass of a gas, the volume of the container, and the temperature. You need to find the pressure using the ideal gas law equation.
02

Convert Mass to Moles

First, determine the number of moles of COâ‚‚. Use the equation: \( n = \frac{m}{M} \). The molecular weight (M) of COâ‚‚ is \( 12.01 + 2 \times 16.00 = 44.01 \ \text{g/mol} \). So, \( n = \frac{1.25 \ \text{g}}{44.01 \ \text{g/mol}} \approx 0.0284 \ \text{mol} \).
03

Convert Temperature to Kelvin

Convert the given temperature in Celsius to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). Therefore, \( T = 22.5 + 273.15 = 295.65 \ \text{K} \).
04

Convert Volume to Liters

Convert the volume from milliliters to liters: \( 750 \ \text{mL} = 0.750 \ \text{L} \).
05

Apply the Ideal Gas Law

Use the ideal gas law equation: \( PV = nRT \). Solving for pressure (P), \( P = \frac{nRT}{V} \). Substitute \( n = 0.0284 \ \text{mol} \), \( R = 0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K} \), \( T = 295.65 \ \text{K} \), \( V = 0.750 \ \text{L} \).
06

Calculate Pressure

Substitute the values into the pressure equation: \( P = \frac{0.0284 \times 0.0821 \times 295.65}{0.750} \). Calculate to obtain \( P \approx 0.921 \ \text{atm} \).

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Key Concepts

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

Pressure Calculation
The concept of pressure calculation in gases can be quite intriguing but straightforward when using the ideal gas law. This law relates pressure, volume, temperature, and the number of moles of a gas. For pressure calculation, the formula is expressed as:\[ P = \frac{nRT}{V} \]Where:
  • \(P\) is the pressure
  • \(n\) is the number of moles of the gas
  • \(R\) is the ideal gas constant (0.0821 L·atm/mol·K)
  • \(T\) is the temperature in Kelvin
  • \(V\) is the volume in liters
To find pressure, you rearrange the ideal gas law equation to solve for \(P\), then substitute the values you have converted: moles of gas, temperature in Kelvin, and volume in liters. This calculation gives you the pressure in atmospheres, a common unit of pressure in chemistry.
By understanding each variable in the equation and knowing how to calculate them correctly, determining the pressure of a gas becomes an easy task.
Mole Conversion
Mole conversion is a fundamental concept in chemistry, especially when working with gases, as it helps in converting mass into a usable quantity for calculations. To convert the mass of a substance into moles, you use the formula:\[ n = \frac{m}{M} \]Where:
  • \(n\) is the number of moles
  • \(m\) is the mass of the substance (in grams)
  • \(M\) is the molar mass (molecular weight) of the substance (in g/mol)
In our exercise with \( \text{CO}_2 \), the molecular weight is calculated by summing the atomic masses of carbon (12.01 g/mol) and oxygen (2 * 16.00 g/mol), resulting in 44.01 g/mol. By dividing the sample mass by this molar mass, you obtain the number of moles, which is a critical step in further calculations involving gases. This standard method is used to bridge the gap between a substance's mass and its chemical amount, helping in precise analytical outcomes.
Temperature Conversion
In chemistry, especially when dealing with gases, it is crucial to express temperature in absolute terms or Kelvin. This ensures consistency and accuracy in calculations. Converting Celsius to Kelvin is straightforward:\[ T(K) = T(°C) + 273.15 \]This conversion is necessary because the Kelvin scale starts at absolute zero, making it ideal for scientific calculations where you often need to multiply or divide temperature values.
In our example, converting \(22.5^{\circ} \text{C}\) to Kelvin gives us \(295.65 \text{K}\). By using Kelvin, you ensure that the temperature values are positive, which is essential for the proportional relationships in the ideal gas law. It is crucial to remember this conversion step as it lays the foundation for accurate pressure, volume, and mole calculations.
Volume Conversion
Volume conversion is an important step when working with the ideal gas law, ensuring that all units are in the proper form. Since the gas constant \(R\) is generally given in L·atm/mol·K, it is necessary to convert the volume of gas from milliliters to liters for calculations. This conversion is simple:\[ V (\text{L}) = V (\text{mL}) \times \frac{1}{1000} \]In the exercise, the \(750 \text{ mL}\) is converted to \(0.750 \text{ L}\) using the conversion factor. This step is crucial because keeping consistent units ensures the mathematical relationships in the ideal gas law are accurate. By converting volumes properly, you eliminate any potential errors in calculations that arise from units inconsistency, ensuring that your final results reflect the correct physical properties of the gas involved.

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Most popular questions from this chapter

The average barometric pressure at an altitude of \(10 \mathrm{km}\) is \(210 \mathrm{mm}\) Hg. Express this pressure in atmospheres, bars, and kilopascals.

If you place \(2.25 \mathrm{g}\) of solid silicon in a \(6.56-\mathrm{L}\). flask that contains \(\mathrm{CH}_{3} \mathrm{Cl}\) with a pressure of \(585 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\) what mass of dimethyldichlorosilane, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g}),\) can be formed? $$\mathrm{Si}(\mathrm{s})+2 \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Si} \mathrm{Cl}_{2}(\mathrm{g})$$ What pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) would you expect in this same flask at \(95^{\circ}\) C on completion of the reaction? (Dimethyldichlorosilane is one starting material used to make silicones, polymeric substances used as lubricants, antistick agents, and water-proofing caulk.)

The hydrocarbon octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) burns to give \(\mathrm{CO}_{2}\) and water vapor: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})+25 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If a \(0.095-\mathrm{g}\) sample of octane burns completely in \(\mathrm{O}_{2},\) what will be the pressure of water vapor in a \(4.75-\mathrm{L}\). Thask at \(30.0^{\circ} \mathrm{C} ?\) If the \(\mathrm{O}_{2}\) gas needed for complete combustion was contained in a \(4.75-\mathrm{L}\). flask at \(22^{\circ} \mathrm{C},\) what would its pressure be?

You have \(3.5 \mathrm{L}\) of \(\mathrm{NO}\) at a temperature of \(22.0^{\circ} \mathrm{C} .\) What volume would the NO occupy at \(37^{\circ} \mathrm{C} ?\) (Assume the pressure is constant.)

Chloroform is a common liquid used in the laboratory. It vaporizes readily. If the pressure of chloroform vapor in a flask is \(195 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C},\) and the density of the vapor is \(1.25 \mathrm{g} / \mathrm{L},\) what is the molar mass of chloroform?

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