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Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 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\(^\circ\)C? (b) What is the force when the temperature of the gas is increased to 100.0\(^\circ\)C?

Short Answer

Expert verified
At 20°C: \( F_1 \approx 8.87 \times 10^4 \, \text{N} \); at 100°C: \( F_2 \approx 1.13 \times 10^5 \, \text{N} \).

Step by step solution

01

Convert Temperature to Kelvin

To use the ideal gas law, we need to convert the temperature from Celsius to Kelvin. The conversion formula is \( T(K) = T(°C) + 273.15 \). For part (a), the temperature is \( T_1 = 20.0 + 273.15 = 293.15 \ K \). For part (b), the temperature is \( T_2 = 100.0 + 273.15 = 373.15 \ K \).
02

Calculate the Volume of the Box

The box is a cube with sides of length 0.300 m. The volume \( V \) of a cube is given by \( V = a^3 \), where \( a \) is the side length. Therefore, \( V = (0.300 \, \text{m})^3 = 0.027 \, \text{m}^3 \).
03

Use the Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant \( R = 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \), and \( T \) is the temperature in Kelvin.
04

Calculate the Pressure at 20.0°C

Substitute the known values into the ideal gas law for \( T_1 = 293.15 \, \text{K} \):\[ P_1 = \frac{nRT_1}{V} = \frac{(3 \, \text{mol})(8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1})(293.15 \, \text{K})}{0.027 \, \text{m}^3} \].Calculate \( P_1 \) to find the pressure.
05

Calculate the Pressure at 100.0°C

For \( T_2 = 373.15 \, \text{K} \), calculate the new pressure \( P_2 \) using the ideal gas law:\[ P_2 = \frac{nRT_2}{V} = \frac{(3 \, \text{mol})(8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1})(373.15 \, \text{K})}{0.027 \, \text{m}^3} \].Calculate \( P_2 \).
06

Calculate the Force on One Side of the Box

The force exerted by the gas on one side of the box is given by \( F = PA \), where \( A = a^2 \) is the area of one side of the cube. Since the side length \( a = 0.300 \, \text{m} \), \( A = (0.300 \, \text{m})^2 \). Calculate the force for both pressures \( P_1 \) and \( P_2 \).

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Key Concepts

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

Temperature Conversion
When dealing with gases, the Ideal Gas Law is a powerful tool for calculating various properties of the gas. However, it requires temperatures to be in Kelvin. To convert from Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
This conversion is vital because Kelvin is the standard unit of measurement in thermodynamic equations. For instance, 20.0°C converts to 293.15 K, and 100.0°C converts to 373.15 K. Always remember to add 273.15 to your Celsius temperature! This ensures your calculations align with the Ideal Gas Law, making your results accurate and reliable.
Volume of a Cube
The volume of a cube is calculated by raising the side length to the power of three. This is because a cube is a three-dimensional shape with equal-length sides.
The formula is simple:
\[ V = a^3 \]
where \( a \) is the side length. For a cube with sides measuring 0.300 m, the volume is:
\[ V = (0.300 \, \text{m})^3 = 0.027 \, \text{m}^3 \]Calculating the volume is crucial in determining how much space the gas molecules have to move around inside the cube. This space influences the pressure and the behavior of the gas inside the container.
Pressure Calculation
Pressure in gases can be calculated using the Ideal Gas Law, which relates pressure, volume, the number of moles, the gas constant, and temperature. The formula is:
\[ PV = nRT \]
To find the pressure \( P \), rearrange the formula:
\[ P = \frac{nRT}{V} \]
Where:
  • \( n \) = number of moles (3 moles in this problem)
  • \( R \) = ideal gas constant \( (8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}) \)
  • \( T \) = temperature in Kelvin
  • \( V \) = volume in cubic meters
By substituting the known values, you can calculate the gas pressure at different temperatures. This in turn helps to find how the pressure changes as temperature varies, deeply impacting the force the gas exerts on the container walls.
Force Exerted by Gas
The force exerted by a gas on the walls of its container can be understood through the relationship between pressure and area. The formula is simple:
\[ F = PA \]
where \( F \) is the force, \( P \) is the pressure, and \( A \) is the area of one side of the container. For a cube, the area of one side is:
\[ A = a^2 \]
In this example, \( a = 0.300 \, \text{m} \) leading to:
\[ A = (0.300 \, \text{m})^2 \]Therefore, you multiply the calculated pressure by this area to find the force on one side. Calculating this for different temperatures helps understand how temperature influences the pressure and thus the force exerted by the gas in the cube. This makes it vital for applications involving gas behavior in confined spaces.

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Most popular questions from this chapter

Modern vacuum pumps make it easy to attain pressures of the order of 10\({^-}{^1}{^3}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of 9.00\(\times\) 10\({^-}{^1}{^4}\) atm and an ordinary temperature of 300.0 K, how many molecules are present in a volume of 1.00 cm\(^3\)? (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

Oxygen (O\(_2\)) has a molar mass of 32.0 g/mol. What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of 300 K; (b) the average value of the square of its speed; (c) the root-mean-square speed; (d) the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 m on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are contained in a vessel of this size at 300 K and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

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