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Chapter 12: Problem 100

Two flasks, each with a volume of \(1.00 \mathrm{L},\) contain \(\mathrm{O}_{2}\) gas with a pressure of \(380 \mathrm{mm}\) Hg. Flask \(\mathrm{A}\) is at \(25^{\circ} \mathrm{C},\) and flask \(\mathrm{B}\) is at \(0^{\circ} \mathrm{C}\). Which flask contains the greater number of \(\mathrm{O}_{2}\) molecules?

Short Answer

Expert verified
Flask B contains more \( \text{O}_2 \) molecules.

Step by step solution

01

Identify Relevant Formula

The ideal gas law is applicable to this problem, which states that \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
02

Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin. For Flask A, the temperature in Kelvin is \( 25 + 273.15 = 298.15 \). For Flask B, it is \( 0 + 273.15 = 273.15 \).
03

Apply the Ideal Gas Law

For each flask, calculate the number of moles \( n \) using the formula \( n = \frac{PV}{RT} \). Assume the pressure \( P = 380 \) mm Hg = \( 380/760 \text{ atm} \), the volume \( V = 1.00 \) L, and \( R = 0.0821 \text{ L atm K}^{-1}\text{ mol}^{-1} \).
04

Calculate Moles in Flask A

Substitute the known values for Flask A: \[ n_A = \frac{(380/760) \times 1.00}{0.0821 \times 298.15} \approx 0.0160 \text{ moles} \].
05

Calculate Moles in Flask B

Substitute the known values for Flask B: \[ n_B = \frac{(380/760) \times 1.00}{0.0821 \times 273.15} \approx 0.0175 \text{ moles} \].
06

Comparison of Number of Moles

Compare the values obtained for \( n_A \) and \( n_B \). Since \( n_B > n_A \), Flask B contains more moles of \( \text{O}_2 \).
07

Conclusion

More moles directly correlate with more molecules, hence Flask B contains the greater number of \( \text{O}_2 \) molecules.

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Key Concepts

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

Moles Calculation
When dealing with gases, understanding how to calculate the number of moles is crucial. The Ideal Gas Law, presented as \( PV = nRT \), allows us to find the number of moles \( n \) of a gas based on known conditions of pressure \( P \), volume \( V \), and temperature \( T \).To perform the moles calculation, rearrange the formula to solve for \( n \):
  • \( n = \frac{PV}{RT} \)
  • **Pressure \( P \)**: Ensure consistency in units; convert mm Hg to atm by dividing by 760, since 1 atm = 760 mm Hg.
  • **Volume \( V \)**: Should be measured in liters (L).
  • **Gas constant \( R \)**: Use the value \( 0.0821 \text{ L atm K}^{-1}\text{ mol}^{-1}\).
  • **Temperature \( T \)**: Must be in Kelvin to match the units of \( R \).
By inserting your values for each variable into the rearranged equation, you can calculate the precise amount of moles of the gas present in your container. This is helpful, as moles directly correlate with the number of molecules present.
Temperature Conversion
When working with gases and using the Ideal Gas Law, it is critical to convert temperatures to Kelvin. The Kelvin scale is an absolute temperature scale starting from absolute zero, which is essential for calculations involving temperature in physics and chemistry.To convert Celsius to Kelvin, you should:
  • Add 273.15 to your Celsius temperature.
For example:
  • A temperature of 25掳C becomes \( 25 + 273.15 = 298.15 \) K.
  • A temperature of 0掳C becomes \( 0 + 273.15 = 273.15 \) K.
Using Kelvin ensures consistency in calculations since the Ideal Gas Law and the gas constant \( R \) utilize this unit. It accommodates the linear relationship between temperature and energy, allowing precise and error-free calculations.
Pressure and Volume
Pressure and Volume are pivotal when applying the Ideal Gas Law. Both must be considered with the appropriate units to ensure an accurate calculation of moles and molecules.**Pressure**- It's important to convert pressure from mm Hg (millimeters of mercury) to atm (atmospheres) for use in the Ideal Gas Law because the standard gas constant \( R \) is often given in terms of atm.- To convert, divide your pressure in mm Hg by 760.**Volume**- Volume should be measured in liters (L) to align with the units of the gas constant \( R \).- Keep in mind that when the container's capacity changes while keeping temperature and pressure constant, the number of molecules and therefore moles will adjust accordingly according to the Ideal Gas Law.Both pressure and volume are inversely related鈥攁s one increases, the other tends to decrease given a constant temperature and quantity of gas. Understanding their interplay is key to solving problems regarding gas behavior and molecular presence.

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Most popular questions from this chapter

Phosphine gas, \(\mathrm{PH}_{3},\) is toxic when it reaches a concentration of \(7 \times 10^{-5} \mathrm{mg} / \mathrm{L} .\) To what pressure does this correspond at \(25^{\circ} \mathrm{C} ?\)

You have two flasks of equal volume. Flask A contains \(\mathrm{H}_{2}\) at \(0^{\circ} \mathrm{C}\) and 1 atm pressure. Flask \(\mathrm{B}\) contains \(\mathrm{CO}_{2}\) gas at \(25^{\circ} \mathrm{C}\) and 2 atm pressure. Compare these two gases with respect to each of the following: (a) average kinetic energy per molecule (b) average molecular velocity (c) number of molecules (d) mass of gas

A \(1.0-\) L flask contains \(10.0 \mathrm{g}\) each of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) (a) Which gas has the greater partial pressure, \(\mathbf{O}_{2}\) or \(\mathbf{C O}_{2}\) or are they the same? (b) Which molecules have the greater average speed, or are they the same? (c) Which molecules have the greater average kinetic energy, or are they the same?

The temperature of the atmosphere on Mars can be as high as \(27^{\circ} \mathrm{C}\) at the equator at noon, and the atmospheric pressure is about \(8 \mathrm{mm}\) Hg. If a spacecraft could collect 10\. \(\mathrm{m}^{3}\) of this atmosphere, compress it to a small volume, and send it back to Earth, how many moles would the sample contain?

Methane is burned in a laboratory Bunsen burner to give \(\mathrm{CO}_{2}\) and water vapor. Methane gas is supplied to the burner at the rate of \(5.0 \mathrm{L} / \mathrm{min}\) (at a temperature of \(28^{\circ} \mathrm{C}\) and a pressure of \(773 \mathrm{mm} \mathrm{Hg}\) ). At what rate must oxygen be supplicd to the burner (at a pressure of \(742 \mathrm{mm} \mathrm{Hg}\) and a temperature of \(26^{\circ} \mathrm{C}\) ) \(?\)
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