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Chapter 15: Problem 6

\(\bullet\) Three moles of an ideal gas are in a rigid cubical box with sides of length 0.200 \(\mathrm{m} .\) (a) What is the force that the gas exerts on each of the six sides of the box when the gas temperature is \(20.0^{\circ} \mathrm{C}\) ? (b) What is the force when the temperature of the gas is increased to \(100.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
At 20.0掳C, the force is 3656 N; at 100.0掳C, it is 4655 N.

Step by step solution

01

Understand the Ideal Gas Law

The ideal gas law is given by the equation \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant \( (8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}) \), and \( T \) is the temperature in Kelvin. This formula relates the pressure, volume, and temperature of an ideal gas.
02

Convert Celsius to Kelvin

To use the ideal gas law, we must convert the given temperatures from Celsius to Kelvin. Use the formula: \( T(\mathrm{K}) = T(\mathrm{C}) + 273.15 \). For \(20.0^{\circ} C\), \( T = 293.15 \) K. For \(100.0^{\circ} C\), \( T = 373.15 \) K.
03

Calculate Volume of the Box

The volume \( V \) of the cubical box can be calculated as \( V = \text{side length}^3 \). With each side of the box measuring 0.200 m, the volume is \( V = (0.200)^3 = 0.008 \mathrm{m}^3 \).
04

Calculate Pressure at 20.0掳C

Using \( PV = nRT \), solve for pressure \( P \): \( P = \frac{nRT}{V} \). Substitute \( n = 3 \ mol \), \( R = 8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} \), \( T = 293.15 \ K \), and \( V = 0.008 \mathrm{m}^3 \). \( P = \frac{3 \times 8.314 \times 293.15}{0.008} = 91400 \mathrm{Pa} \).
05

Calculate Force on One Side at 20.0掳C

The force on one side of the box is given by \( F = P \cdot A \), where \( A = \text{side length}^2 \). For a side length of 0.200 m, \( A = 0.200^2 = 0.040 \mathrm{m}^2 \). So, \( F = 91400 \mathrm{Pa} \times 0.040 \mathrm{m}^2 = 3656 \mathrm{N} \).
06

Calculate Pressure at 100.0掳C

Repeat Step 4 with the new temperature \( T = 373.15 \ K \). \( P = \frac{3 \times 8.314 \times 373.15}{0.008} = 116370 \mathrm{Pa} \).
07

Calculate Force on One Side at 100.0掳C

Using the pressure from Step 6, calculate the force: \( F = 116370 \mathrm{Pa} \times 0.040 \mathrm{m}^2 = 4655 \mathrm{N} \).

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Key Concepts

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

Temperature Conversion
Converting temperature from Celsius to Kelvin is crucial when working with the Ideal Gas Law. This is because the law uses the Kelvin scale for temperature, which starts at absolute zero. The conversion is simple:
  • Add 273.15 to the Celsius temperature.
  • This results in the Kelvin temperature.
For instance, if the temperature of the gas is 20.0掳C, converting it gives 293.15 K. Similarly, a temperature of 100.0掳C would convert to 373.15 K. This conversion ensures accurate calculations when applying the Ideal Gas Law.
Pressure Calculation
Calculating pressure in an ideal gas is an exciting aspect of physics. The Ideal Gas Law, represented as \( PV = nRT \), helps us find pressure \( P \) when the other variables are known.Take these steps to calculate pressure:
  • Identify the number of moles \( n \) of gas present.
  • Use the universal gas constant \( R = 8.314 \, \mathrm{J/mol} \, \mathrm{K} \).
  • Estimate the volume \( V \) and convert the temperature to Kelvin \( T \).
To find the pressure, rearrange the Ideal Gas Law to \( P = \frac{nRT}{V} \). By plugging in the values, you can calculate the accurate pressure exerted by the gas on its container.
Volume Calculation
When dealing with gases, understanding the concept of volume is necessary. For a cubical box, volume \( V \) is calculated using the formula \( V = \text{side length}^3 \).
To find the volume of the box:
  • Measure the length of one side of the cube.
  • Cube this value (multiply the number by itself three times).
For example, a box with sides of 0.200 m will have a volume of \( 0.200^3 = 0.008 \, \mathrm{m}^3 \). Volume is key in the Ideal Gas Law where it influences the pressure according to the equation \( PV = nRT \).
Force Calculation
Force exerted by a gas on the walls of its container can be derived from the pressure. Given the pressure \( P \) and the surface area \( A \) of one wall, use the formula \( F = P \times A \) to compute the force.
  • First, calculate the area of one side of the box; for a square, \( A = \text{side length}^2 \).
  • Multiply the pressure by this area to obtain the force.
Taking a box with side length 0.200 m as an example, the area is \( 0.040 \, \mathrm{m}^2 \). If pressure at 20掳C is 91400 Pa, then the force is \( 91400 \, ext{Pa} \times 0.040 \, ext{m}^2 = 3656 \, ext{N} \). Understanding this calculation is essential for predicting how changes in temperature and pressure affect the gas.

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Most popular questions from this chapter

\(\bullet\) Suppose some insects have speeds of \(10.00 \mathrm{m} / \mathrm{s}, 8.00 \mathrm{m} / \mathrm{s},\) \(7.00 \mathrm{m} / \mathrm{s},\) and 2.00 \(\mathrm{m} / \mathrm{s} .\) Find (a) the rms speed of these critters and (b) their average speed.

\(\bullet\) An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

Near room temperature, how does the internal energy of one mole of a diatomic ideal gas compare to that of one mole of a monatomic ideal gas? A. They have the same internal energy. B. The diatomic gas has 2 times as much internal energy as the monatomic gas. C. The diatomic gas has 2\(/ 3\) times as much internal energy as the monatomic gas. D. The diatomic gas has 3\(/ 2\) times as much internal energy as the monatomic gas. E. The diatomic gas has 5\(/ 3\) times as much internal energy as the monatomic gas.

\(\bullet$$\bullet\) The atmosphere of the planet Mars is 95.3\(\%\) carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and about 0.03\(\%\) water vapor. The atmospheric pressure is only about 600 \(\mathrm{Pa}\) , and the surface temperature varies from \(-30^{\circ} \mathrm{C}\) to \(-100^{\circ} \mathrm{C}\) . The polar ice caps contain both \(\mathrm{CO}_{2}\) ice and water ice. Could there be liquid \(\mathrm{CO}_{2}\) on the surface of Mars? Could there be liquid water? Why or why not?

\(\bullet$$\bullet\) Heat \(Q\) flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?
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