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Suppose 2.00 moles of an ideal gas is enclosed in a container of volume \(1.00 \cdot 10^{-4} \mathrm{~m}^{3}\). The container is then placed in a furnace and raised to a temperature of \(400 . \mathrm{K}\). What is the final pressure of the gas?

Short Answer

Expert verified
Answer: The pressure of the ideal gas is \(6.656 \cdot 10^7 \mathrm{~Pa}\).

Step by step solution

01

Identify the given values

We are given the following values: Number of moles (n): 2.00 moles Volume (V): \(1.00 \cdot 10^{-4} \mathrm{~m}^{3}\) Temperature (T): 400 K
02

Write down the Ideal Gas Law formula

The Ideal Gas Law formula is given by: \(PV=nRT\) Our goal is to calculate the pressure (P).
03

Use the given values and the Ideal Gas Law formula to solve for the pressure

We can rearrange the Ideal Gas Law formula to solve for the pressure by dividing both sides of the equation by the volume (V): \(P = \frac{nRT}{V}\) Now, let's plug in the given values into the formula: \(P = \frac{2.00 \cdot 8.31 \cdot 400}{1.00 \cdot 10^{-4}}\)
04

Calculate the pressure

Now, let's perform the arithmetic to find the pressure: \(P = \frac{2.00 \cdot 8.31 \cdot 400}{1.00 \cdot 10^{-4}} = \frac{6656}{1.00 \cdot 10^{-4}} = 6.656 \cdot 10^7 \mathrm{~Pa}\) The final pressure of the gas is \(6.656 \cdot 10^7 \mathrm{~Pa}\).

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Key Concepts

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

Thermodynamics and the Ideal Gas Law
When studying thermodynamics, the behavior of gas molecules under different conditions of temperature, pressure, and volume often comes into focus. A central piece of this puzzle is the Ideal Gas Law, represented by the formula \(PV=nRT\). This elegant equation encapsulates how these properties of a gas are interconnected.

In the context of our exercise, we are observing what happens to an ideal gas when it is heated inside a furnace. As per the kinetic molecular theory, increasing the temperature results in an increase in the kinetic energy of the gas molecules. Consequently, the molecules move more rapidly and collide with the container's walls more frequently and forcefully, leading to an increase in pressure, assuming the volume remains constant.

The Ideal Gas Law is particularly useful in this scenario because it allows for the calculation of one missing property (like pressure) when the others (volume, moles of gas, and temperature) are known. By understanding this fundamental relationship, you have tackled one of the cornerstones of thermodynamics.
Pressure Calculation Using the Ideal Gas Law
The pressure of a gas is a measure of the force exerted by the gas molecules against the surface area of their container. Calculating this pressure can seem challenging, but with the Ideal Gas Law, it becomes a manageable task.

To calculate the pressure of an ideal gas, the formula \(P = \frac{nRT}{V}\) is used. Here, \(P\) stands for pressure, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, \(T\) is the absolute temperature in Kelvins, and \(V\) is the volume. This rearrangement of the Ideal Gas Law formula is straightforward and explicitly solves for pressure.

In our specific exercise, after substituting the given values into this equation, we find that the pressure of the gas increases significantly upon being heated to 400 K. It's important to remember that R, the universal gas constant, has a fixed value of approximately 8.31 \(\frac{J}{mol \cdot K}\) and it serves as a bridge between the macroscopic and microscopic properties of the gas.
Understanding Moles of Gas in the Ideal Gas Law
The 'mole' is a fundamental concept in both chemistry and physics, serving as a unit for measuring the amount of substance. In terms of gases, a mole can be visualized as a specific number of molecules, typically Avogadro's number \(6.022 \times 10^{23}\) molecules.

The number of moles in a gas is directly proportional to the number of molecules present. This directly affects the pressure and volume of the gas, which can be predicted using the Ideal Gas Law. In our exercise, 2.00 moles of the gas are being considered. This directly tells us how many molecules are buzzing around and bumping into each other and the walls of the container.

It is fascinating that through the simple notation 'n' for moles in the Ideal Gas Law, we can equate a macroscopic property like pressure to the microscopic reality of molecular behavior. Understanding moles helps us to calculate with precision how changes in conditions like temperature affect pressure, as neatly demonstrated in our ideal gas scenario.

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Most popular questions from this chapter

At Party City, you purchase a helium-filled balloon with a diameter of \(40.0 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\) and at \(1.00 \mathrm{~atm} .\). a) How many helium atoms are inside the balloon? b) What is the average kinetic energy of the atoms? c) What is the root-mean-square speed of the atoms?

How much heat is added to the system when \(2.00 \mathrm{~kJ}\) of work is performed by an ideal gas in an isothermal process? Give a reason for your answer.

Interstellar space far from any stars is usually filled with atomic hydrogen \((\mathrm{H})\) at a density of \(1.00 \mathrm{atom} / \mathrm{cm}^{3}\) and a very low temperature of \(2.73 \mathrm{~K}\) a) Determine the pressure in interstellar space. b) What is the root-mean-square speed of the atoms? c) What would be the edge length of a cube that would contain atoms with a total of 1.00 J of energy? 19.46 a) What is the root-mean-square speed for a collection of helium- 4 atoms at \(300 .\) K? b) What is the root-mean-square speed for a collection of helium-3 atoms at \(300 . \mathrm{K} ?\)

A closed auditorium of volume \(2.50 \cdot 10^{4} \mathrm{~m}^{3}\) is filled with 2000 people at the beginning of a show, and the air in the space is at a temperature of \(293 \mathrm{~K}\) and a pressure of \(1.013 \cdot 10^{5} \mathrm{~Pa}\). If there were no ventilation, by how much would the temperature of the air rise during the 2.00 -h show if each person metabolizes at a rate of 70.0 W?

In a diesel engine, the fuel-air mixture is compressed rapidly. As a result, the temperature rises to the spontaneous combustion temperature for the fuel, causing the fuel to ignite. How is the rise in temperature possible, given the fact that the compression occurs so rapidly that there is not enough time for a significant amount of heat to flow into or out of the fuel-air mixture?

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