/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Three moles of an ideal gas are ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Three moles of an ideal gas are in a rigid cubical box with sides of length \(0.300 \mathrm{~m}\). (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 exerted by the gas on each side of the box at \(20.0^{\circ} \mathrm{C}\) (@293.15K) and \(100.0^{\circ} \mathrm{C}\) (@373.15K) can be determined by applying the ideal gas law to find the pressure of the gas at those temperatures. The force is then found by multiplying the pressure by the area of a side of the box.

Step by step solution

01

Convert Temperatures to Kelvin

Since the ideal gas law requires temperatures to be in Kelvin, it is necessary to convert the given temperatures, which are in degrees Celsius, to Kelvin. The formula for this conversion is \(T(K) = T(°C) + 273.15\). Therefore, the temperature at \(20.0^{\circ} \mathrm{C}\) equals \(293.15 \mathrm{K}\), and at \(100.0^{\circ} \mathrm{C}\) equals \(373.15 \mathrm{K}\).
02

Calculate the Volume

The volume of a cubical box is given by \(V = L^3\) where \(L\) is the length of a side. Since the length of each side is \(0.300 \mathrm{~m}\), the volume of the box is \(V = (0.300 \mathrm{~m})^3 = 0.027 \mathrm{~m}^3\).
03

Determine the pressure

Having the volume, the amount of gas (in moles) and the temperatures (in Kelvin), we can use the ideal gas law to calculate the pressure in both cases. The ideal gas constant \(R = 8.31 \mathrm{~J} / (\mathrm{K} \cdot \mathrm{mol})\) also needs to be taken into account. The pressure at \(20.0^{\circ} \mathrm{C}\) (\(293.15 \mathrm{K}\)) is given by \(P_1 = \frac{nRT_1}{V}\) and at \(100.0^{\circ} \mathrm{C}\) (\(373.15 \mathrm{K}\)) by \(P_2 = \frac{nRT_2}{V}\). By substituting the appropriate values, the pressures can be found.
04

Determine the Force

We can now determine the force exerted by the gas on each side of the box. The force is given by \(F = P \cdot A\), where \(A = L^2\), the area of each side of the box. Substituting the pressures obtained in Step 3 and calculating, the force at \(20.0^{\circ} \mathrm{C}\) and at \(100.0^{\circ} \mathrm{C}\) can be determined.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics
Thermodynamics is the study of the relationships between heat, work, and energy. One of its key principles is the behavior of gases, which can be described using the ideal gas law. This law is particularly important in understanding how temperature, pressure, and volume interact in a system with a fixed amount of gas.

The ideal gas law is usually expressed as \(PV=nRT\), where \(P\) stands for pressure, \(V\) is the volume, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. For the three moles of gas in our rigid cubical box problem, the ideal gas law can be applied to determine how the gas will exert force on the box’s sides when the temperature changes.

This law explains that if the volume of the gas and the number of moles remain constant but the temperature increases, the pressure inside the box will also increase. Consequently, the force exerted by the gas on each side of the box will rise. Understanding these relationships is fundamental in predicting the behavior of gases in real-life applications such as engines, refrigerators, and even weather systems.
Pressure Calculation
Pressure calculation using the ideal gas law is central to solving problems involving gases in enclosed spaces. Pressure is understood as the force applied per unit area and can be affected by changes in temperature and volume in a gas system.

From the given problem, after converting the given temperatures from Celsius to Kelvin, we calculated the volume of the cubical box. We then used the ideal gas law formula to find the pressure exerted by the gas at different temperatures. The force that the gas applies on any side of the box is directly related to the pressure within the box; the higher the pressure, the greater the force. For instance, if the temperature increases, the pressure also increases assuming that the volume cannot change since the box is rigid. This illustration of pressure calculation shows how a seemingly abstract concept has direct, observable physical effects.
Temperature Conversion
Temperature conversion is essential in thermodynamics and particularly when using the ideal gas law, as it requires temperature to be measured in Kelvin rather than Celsius or Fahrenheit. To solve problems involving the ideal gas law, temperatures in Celsius must be converted to Kelvin using the conversion formula \(T(K) = T(°C) + 273.15\).

In the textbook exercise, the temperatures provided were in Celsius. To apply the ideal gas law, we converted a room-temperature of \(20.0^{\textdegree C}\) to \(293.15 \text{K}\), and a higher temperature of \(100.0^{\textdegree C}\) to \(373.15 \text{K}\). These conversions are crucial for obtaining accurate results when calculating pressure changes due to temperature variations inside the box. It's important for students to understand this conversion process because any error made in converting temperatures can lead to incorrect results, potentially misunderstanding a gas's behavior under changing conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It is possible to make crystalline solids that are only one layer of atoms thick. Such "two-dimensional" crystals can be created by depositing atoms on a very flat surface. (a) If the atoms in such a twodimensional crystal can move only within the plane of the crystal, what will be its molar heat capacity near room temperature? Give your answer as a multiple of \(R\) and in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\). (b) At very low temperatures, will the molar heat capacity of a two-dimensional crystal be greater than, less than, or equal to the result you found in part (a)? Explain why.

A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height \(h\) has the value \(3.50 \mathrm{~m}\), the absolute pressure \(p\) of the compressed air is \(4.20 \times 10^{5} \mathrm{~Pa}\). Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5} \mathrm{~Pa}\). (a) What is the speed with which water flows out of the hose when \(h=3.50 \mathrm{~m} ?\) (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h=3.00 \mathrm{~m}\) and for \(h=2.00 \mathrm{~m} .\) (c) At what value of \(h\) does the flow stop?

How many moles are in a \(1.00 \mathrm{~kg}\) bottle of water? How many molecules? The molar mass of water is \(18.0 \mathrm{~g} / \mathrm{mol} .\)

A flask with a volume of \(1.50 \mathrm{~L}\), provided with a stopcock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at \(300 \mathrm{~K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{~Pa}\right) .\) The molar mass of ethane is \(30.1 \mathrm{~g} / \mathrm{mol} .\) The system is warmed to a temperature of \(550 \mathrm{~K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

A balloon of volume \(750 \mathrm{~m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of \(1.90 \mathrm{~m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{~Pa}\), how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at \(15.0^{\circ} \mathrm{C} ?\) The molar mass of hydrogen \(\left(\mathrm{H}_{2}\right)\) is \(2.02 \mathrm{~g} / \mathrm{mol} .\) The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is \(1.23 \mathrm{~kg} / \mathrm{m}^{3} .\) See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass \(4.00 \mathrm{~g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.