Ideal Gas Law

\[P = \frac{\rho k_B T}{m_\text{average}}\quad\text{Eq. (9.6)}\]

Note: If you feel uncomfortable with working with fractions or variables, you should review the following tutorials before working through this tutorial:

Sections

Introduction

Note: If you haven’t reviewed the gas particle speed tutorial, it may help to take a look at that page before proceeding with this tutorial.

\The ideal gas law is absolutely fundamental in understanding the interiors of stars and the atmospheres of planets. Although it is only an approximation to a more sophisticated and accurate expression of pressure, the approximation is very good for many situations. Times when the ideal gas “law” doesn’t work very well, or fails completely, when the gas becomes very dense. These environments can be very extreme, and may lead to spectacular outcomes. We will discuss these cases in Chapters 16 and 17, but for now the ideal gas “law” is perfectly adequate for our needs. Putting quotes around “law” now emphasizes that it is not truly exact in all cases, just like Newton’s “law” of gravity fails in extreme cases, which led to the need for Einstein’s general theory of relativity. Our understanding of nature continues to evolve.

The ideal gas law contains the fundamental constant (\(k_B\)), temperature (\(T\)), and the ratio of mass density to average particle mass (\(\rho/m_\text{average}\)). In the gas particle speed tutorial it is pointed out that particles move faster with increasing temperature and particles that are more massive move more slowly. The ideal gas law describes the gas pressure produced by particles banging into, and bouncing off, one another. The faster the particles are moving (the higher \(T\) is), the greater their change in momentum and the more frequent the collisions. A higher temperature also means that each particle has more kinetic energy. For example, think about what it would be like to get hit by a baseball traveling with a speed of 1 m/s compared to one moving at 30 m/s. You would definitely get pushed harder by the 30 m/s baseball. But the 30 m/s baseball would reach you from the same distance in 1/30 the time. With the same number of particles flying around 30 times faster you would get hit on average 30 times more often in the same amount of time. The number of particles that are flying around in a given volume of gas is given by the mass of all the particles in the volume divided by the average mass of each individual particle in the gas. This is why the ratio \(\rho/m_\text{average}\) is in the ideal gas law along with temperature \(T\).

Having a picture in your mind’s eye of what is happening physically helps you to understand what the mathematics is telling you. Memorization of symbols, or even just looking at them in front of you is not the same as understanding. You should always strive to make sense of why the equations are the way they are.

Examples

  1. An indestructible box contains a gas that is composed entirely of hydrogen atoms (don’t try this at home – think Hindenburg). If the contents of the box is originally at a temperature of \(\text{3000 K}\) and is heated to a temperature of \(\text{6000 K}\), does the gas pressure inside the box increase, decrease, or remain unchanged? If the pressure does change, by what factor does it change?

    Because the particles in the gas don’t change in number or average mass, \(\rho/m_\text{average}\) doesn’t change either. However, increasing the temperature \(T\) means that the gas pressure, \(P\), must increase by the same relative amount. Since the temperature increased by a factor of 2, the pressure also increased by a factor of 2.
  2. If the temperature of the gas in Example 1 dropped from \(\text{3000 K}\) to \(\text{500 K}\) what effect did the temperature change have on the pressure inside the box?

    Since the temperature of the contents decreased by a factor of 6 (\(\text{3000 K}/\text{500 K} = \text{30}/\text{5} = 6\)), the pressure in the gas also decreased by a factor of 6.
  3. Suppose that exactly 1/2 of the hydrogen atoms escape from the box in Example 1, but the initial temperature of 3000 K remains unchanged. What happens to the gas pressure inside the box?

    With 1/2 as many hydrogen atoms in the box, the mass density, \(\rho\), must have decreased by a factor of two while the average mass per particle, \(m_\text{average}\), stayed the same (there are still hydrogen atoms in the box, just one-half as many of them). As a result, the pressure decreased by a factor of 2; the pressure is 1/2 of what it was before the hydrogen atoms escaped.
  4. The original box in Example 1 is replaced by an identical box containing the same amount of mass, but now the particles inside are helium atoms that are four times more massive than hydrogen atoms (this box would be much safer to experiment with by the way!). The temperature inside the new box is also the same as the original box (\(\text{3000 K}\)). Compare the gas pressure inside the original box with the pressure inside the helium-filled box.

    Since the two boxes are identical (not including the contents), they have identical volumes inside. Since they also contain the same amount of mass and are at the same temperature, the only thing that differs is that the individual particles in the helium-filled box are four times more massive but there are one-fourth as many of them. Since \(\rho\) is the same in both boxes, the helium-filled box would have 1/4 the pressure of the hydrogen-filled box (\(m_\text{average}\) is four times greater but \(m_\text{average}\) is in the denominator of the ideal gas law).

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers to the problems are given below)

  1. A cylinder contains a gas at a temperature of \(\text{300 K}\). The volume of the cylinder increases by a factor of five while the temperature remains at \(\text{300 K}\) and no particles escape or are added to the cylinder. What happens to the gas pressure inside the cylinder?
  2. How would the temperature inside the cylinder from Exercise 1 need to change in order to return the pressure in the cylinder to its original value after the volume has increased by a factor of five?
  3. A gas of pure helium in a box has a temperature of \(\text{300 K}\). Gas in another, identical box, contains the same number of nitrogen molecules, where nitrogen molecules are seven times more massive than helium atoms. What would the temperature need to be in the box containing nitrogen molecules in order for the gas pressures in the two boxes to be the same?
  4. Two identical boxes contain identical particles, but box A contains three times as many of the particles as box B. If both boxes have the same gas pressure, which box has the higher temperature, and by what factor?
Answers
  1. The gas pressure decreases by a factor of five.
  2. The temperature of the gas would need to increase by a factor of five to a value of \(\text{1500 K}\).
  3. The temperature would need to be the same. This is because the mass density \(\rho\) would increase by a factor of seven (the same number of particles but the mass of each particle is seven times greater) and the average mass of each particle \(m_\text{average}\) is also seven times greater, which implies that \(\rho/m_\text{average}\) is unchanged.
  4. Since box A contains three times as many identical particles, \(\rho_A\) must be three times greater than \(\rho_B\). As a result, the temperature of the gas in box B must be a factor of three times greater than box A so that \(\rho_A T_A\) equals \(\rho_B T_B\).
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