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Chapter 3: Problem 14

. Under what circumstances is the following statement correct? Equal molar amounts of two different gases at the same temperature, placed in containers of equal volume, have the same pressure.

Short Answer

Expert verified
Equal pressures when molar amount, temperature, and volume are the same.

Step by step solution

01

- Understand the Ideal Gas Law

The Ideal Gas Law states: \[ PV = nRT \]Where:- \(P\) is the pressure,- \(V\) is the volume,- \(n\) is the number of moles,- \(R\) is the universal gas constant,- \(T\) is the temperature.
02

- Analyzing the Given Conditions

Given that the gases have equal molar amounts (i.e., the same number of moles, \(n\)) and are at the same temperature (\(T\)). They are also placed in containers of equal volume (\(V\)).
03

- Applying the Ideal Gas Law to Each Container

For the first gas, the pressure can be expressed as:\[ P_1V = nRT \]For the second gas, the pressure can be expressed as:\[ P_2V = nRT \]Since \(V\), \(n\), \(R\), and \(T\) are the same for both gases, the pressures \(P_1\) and \(P_2\) must also be equal.
04

- Conclusion

Under the conditions that all other variables (molar amount, temperature, volume, and the gas constant) are the same, the pressures of the two gases will be equal.

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Key Concepts

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

Understanding Pressure
Pressure is a force exerted by gas particles when they collide with the walls of their container. In the context of the Ideal Gas Law, it's denoted by the symbol \(P\). Pressure is crucial because it tells us how much force gas particles are applying. In daily life, we measure pressure in different units like atmospheres (atm), pascals (Pa), or pounds per square inch (psi). Higher pressure means particles collide more frequently and forcefully against the container walls.
Defining Volume
Volume (\(V\)) is the amount of space a gas occupies. In the Ideal Gas Law, volume is crucial as it helps to determine how much space gas particles have to move around. The volume of a gas can be measured in liters (L), cubic meters (m鲁), or cubic centimeters (cm鲁). Different volumes under constant temperature and pressure will hold different amounts of gas. In our exercise, the volume is the same for both gases, meaning they have the same amount of space to move around.
Exploring Moles
A mole (\(n\)) is a measurement of the amount of substance and is part of the Ideal Gas Law. One mole contains approximately 6.022 \times 10^{23} particles (Avogadro's number). It's a way to count atoms or molecules. In our scenario, equal molar amounts mean both containers have the same number of gas molecules. This is why the pressure relates directly to the number of moles when temperature and volume remain constant.
The Universal Gas Constant
The universal gas constant (\(R\)) is a constant that appears in the Ideal Gas Law. Its value is approximately 8.314 J/(mol路K). \(R\) provides a relationship factor between the pressure, volume, moles, and temperature of a gas. It helps unify these variables into one equation. Different units of \(R\) are used depending on the units of pressure and volume in a given context. For instance, it's essential when calculating how gases behave under varying conditions.
Importance of Temperature
Temperature (\(T\)) is a measure of the average kinetic energy of gas particles. In the Ideal Gas Law, it's expressed in Kelvin. It directly influences the behavior of gases. Higher temperatures mean gas particles move more energetically, increasing pressure if volume and the number of moles remain constant. In the exercise, both gases are at the same temperature, ensuring they have equivalent kinetic energy. This consistency helps in concluding that pressures remain equal when other factors are constant.

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Most popular questions from this chapter

A rigid tank initially contains \(3 \mathrm{~kg}\) of air at \(500 \mathrm{kPa}, 290 \mathrm{~K}\). The tank is connected by a valve to a piston-cylinder assembly oriented vertically and containing \(0.05 \mathrm{~m}^{3}\) of air initially at \(200 \mathrm{kPa}, 290 \mathrm{~K}\). Although the valve is closed, a slow leak allows air to flow into the cylinder until the tank pressure falls to \(200 \mathrm{kPa}\). The weight of the piston and the pressure of the atmosphere maintain a constant pressure of \(200 \mathrm{kPa}\) in the cylinder; and owing to heat transfer, the temperature stays constant at \(290 \mathrm{~K}\). For the air, determine the total amount of energy transfer by work and by heat, each in kJ. Assume ideal gas behavior.

Compare the densities, in \(\mathrm{kg} / \mathrm{m}^{3}\), of helium and air, each at \(300 \mathrm{~K}, 100 \mathrm{kPa}\). Assume ideal gas behavior.

A system consisting of \(2 \mathrm{~kg}\) of ammonia undergoes a cycle composed of the following processes: Process 1-2: constant volume from \(p_{1}=10\) bar, \(x_{1}=0.6\) to saturated vapor Process 2-3: constant temperature to \(p_{3}=p_{1}, Q_{23}=+228 \mathrm{~kJ}\) Process 3-1: constant pressure Sketch the cycle on \(p-v\) and \(T-v\) diagrams. Neglecting kinetic and potential energy effects, determine the net work for the cycle and the heat transfer for each process, all in \(\mathrm{kJ}\).

Two kilograms of Refrigerant \(134 a\), initially at 2 bar and occupying a volume of \(0.12 \mathrm{~m}^{3}\), undergoes a process at constant pressure until the volume has doubled. Kinetic and potential energy effects are negligible. Determine the work and heat transfer for the process, each in \(\mathrm{kJ}\).

A gas is confined to one side of a rigid, insulated container divided by a partition. The other side is initially evacuated. The following data are known for the initial state of the gas: \(p_{1}=5\) bar, \(T_{1}=500 \mathrm{~K}\), and \(V_{1}=0.2 \mathrm{~m}^{3}\). When the partition is removed, the gas expands to fill the entire container, which has a total volume of \(0.5 \mathrm{~m}^{3}\). Assuming ideal gas behavior, determine the final pressure, in bar.
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