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Chapter 18: Problem 16

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
The force at 20掳C is approx. 36,609.365 N; at 100掳C, it is approx. 46,636.133 N.

Step by step solution

01

Understand the Problem

We need to find the force the gas exerts on each side of the box at given temperatures and using the ideal gas law.
02

Apply the Ideal Gas Law

The ideal gas law is \[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant \((8.314 \, J/(mol \cdot K))\), and \( T \) is the temperature in Kelvin. The volume \( V \) of the box is given by \( V = 0.2^3 = 0.008 \, m^3 \).
03

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \( T(K) = T(掳C) + 273.15 \). - At \( 20.0^{\circ} \mathrm{C} \), \( T = 293.15 \, K \).- At \( 100.0^{\circ} \mathrm{C} \), \( T = 373.15 \, K \).
04

Calculate Pressure at 20.0掳C

Substitute the values into the ideal gas law to solve for pressure \( P \) at \( T = 293.15 \, 掳K \):\[ P = \frac{nRT}{V} = \frac{(3 \, \text{moles})(8.314 \, J/(mol \cdot K))(293.15 \, K)}{0.008 \, m^3} \]\[ P \approx 915,234.125 \, N/m^2 \].
05

Calculate Force on Each Side at 20.0掳C

Use the formula \( F = PA \) where \( A \) is the area of one side of the box, which is \( 0.2^2 = 0.04 \, m^2 \). Thus, the force is:\[ F = 915,234.125 \, N/m^2 \times 0.04 \, m^2 \]\[ F \approx 36,609.365 \, N \].
06

Calculate Pressure at 100.0掳C

Use the same steps to find pressure at \( T = 373.15 \, K \):\[ P = \frac{(3)(8.314)(373.15)}{0.008} \] \[ P \approx 1,165,903.3125 \, N/m^2 \].
07

Calculate Force on Each Side at 100.0掳C

Calculate the force at the new pressure:\[ F = 1,165,903.3125 \, N/m^2 \times 0.04 \, m^2 \]\[ F \approx 46,636.1325 \, N \].

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Key Concepts

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

Pressure Calculation
When calculating the pressure exerted by a gas, we rely on the ideal gas law: \[ PV = nRT \] Here, \( P \) denotes pressure, \( V \) is the volume, \( n \) represents the number of moles, \( R \) is the ideal gas constant, and \( T \) refers to temperature in Kelvin.
In our example, we want to find the pressure the gas exerts within a rigid cubical box with known dimensions. The key steps to calculate this pressure include:
  • Determining the volume of the cube using the formula \( V = \, {\text{side length}}^3 \).
  • Substituting the known values into the ideal gas law to solve for pressure \( P \).
As temperature or the number of moles changes, pressure will adjust according to the ideal gas equation.
Temperature Conversion
Temperature conversion is crucial because the ideal gas law requires temperatures to be in Kelvin rather than Celsius.
The conversion process involves a simple formula: \[ T(K) = T(掳C) + 273.15 \] This formula adds 273.15 to the Celsius temperature to convert it to Kelvin.
In the given problem, the temperatures provided are:\( 20.0^{\circ} \mathrm{C} \) and \( 100.0^{\circ} \mathrm{C} \).
  • Converting \( 20.0^{\circ} \mathrm{C} \) to Kelvin: \( T = 20 + 273.15 = 293.15 \, K \).
  • Converting \( 100.0^{\circ} \mathrm{C} \) to Kelvin: \( T = 100 + 273.15 = 373.15 \, K \).
This conversion ensures that our calculations using the ideal gas law are consistent and accurate.
Force Exerted by Gas
Once we find the pressure, next is to determine the force exerted by the gas on the box walls. This force can be calculated by using the relation:\[ F = PA \] where \( F \) is the force, \( P \) is the pressure, and \( A \) is the area of the surface on which the force is acting.
In this scenario, \( A \) is the area of each side of the cube:
\( A = 0.2^2 = 0.04 \, m^2 \).
  • For a pressure of \( 915,234.125 \, N/m^2 \) at \( 20.0^{\circ} \mathrm{C} \), the force is \( F \approx 36,609.365 \, N \).
  • For a pressure of \( 1,165,903.3125 \, N/m^2 \) at \( 100.0^{\circ} \mathrm{C} \), the force is \( F \approx 46,636.1325 \, N \).
These calculations show how increased temperature results in higher pressure and, subsequently, more force exerted by the gas.
Volume of a Cube
Calculating the volume of a cube is a straightforward task, but very important for applying the ideal gas law. The volume \( V \) of a cube is found by raising the side length to the third power:\[ V = l^3 \] where \( l \) is the length of a side of the cube.
For a cube with a side length of \( 0.200 \, m \), the volume would be:\[ V = 0.2^3 = 0.008 \, m^3 \]
Understanding how to determine the cube's volume is essential, as it directly influences the application of the ideal gas law. The calculated volume is used as the \( V \) in the ideal gas equation, ensuring the accuracy of the pressure and force results.

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Most popular questions from this chapter

(a) Compute the specific heat capacity at constant volume of nitrogen \(\left(\mathrm{N}_{2}\right)\) gas, and compare with the specific heat capacity of liquid water. The molar mass of \(\mathrm{N}_{2}\) is 28.0 \(\mathrm{g} / \mathrm{mol}\) . (b) You warm 1.00 \(\mathrm{kg}\) of water at a constant volume of 1.00 \(\mathrm{L}\) from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\) in a kettle. For the same amount of beat, how many kilograms of \(20.0^{\circ} \mathrm{C}\) air would you be able to warm to \(30.0^{\circ} \mathrm{C} ?\) What volume (in liters) would this air occupy at \(20.0^{\circ} \mathrm{C}\) and a pressure of 1.00 atm? Make the simplifying assumption that air is 100\(\% \mathrm{N}_{2}\) .

Helium gas with a volume of 2.60 \(\mathrm{L}\) , under a pressure of 1.30 atm and at a temperanure of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol.}\)

(a) Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol}\) . What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K} ?\) (b) What is the average value of the square at a of its speed? (c) What is the root-mean-square speed? (d) What is 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 \(\mathrm{m}\) 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 aver- age force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 \(\mathrm{atm} ?\) (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) 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?

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 \(\mathrm{m}\) . The temperature at the surface was \(27.0^{\circ} \mathrm{C},\) and at the bottom it was \(7.0^{\circ} \mathrm{C}\) . The density of seawater is 1030 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 \(\mathrm{m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell. \((b)\) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 \(\mathrm{m}^{3}\) of air at pressure of 3.40 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 \(\mathrm{m}^{3} .\) If the temperature remains constant, what is the final value of the pressure?
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