Ideal Gas Law
Task
A certain number of Xenon gas molecules are placed in a container at room temperature. If $V$ is the volume of the container and $P(V)$ is the pressure exerted on the container by the Xenon molecules, a model predicts that $$P(V) = \frac{40}{2V-1}-\left(\frac{4}{V}\right)^2$$ for all $V>\frac{1}{2}$. Here the units for volume are liters and the units for pressure are atmospheres.Â
- Sketch a graph of $P$.
- Using the graph, approximate the volume for which the pressure is 10 atmospheres.
IM Commentary
The goal of this task is to interpret the graph of a rational function and use the graph to approximate when the function takes a given value. The first two parts of the question focus student attention on the meaning of the function within the context of pressure and volume. Students must identify which volumes make sense and then understand that the meaningful inputs of the function are where $V \gt \frac{1}{2}$. The task presents a small number of questions about this context, but instructors looking to probe more deeply have a number of possibilities at their disposal. For example, after they have sketched the graph in part (b), the teacher may wish to ask students what the vertical asymptote at $V = \frac{1}{2}$ means in terms of the context, or to explain the meaning of the horizontal asymptote as the volume grows larger and larger. While students should be able to produce a qualitatively correct picture of the graph in part (b), technology is recommended to get a reasonably accurate answer to (c). Using the graph to make estimates is appropriate in this situation as trying to find out the exact volume for which the pressure will be 10 atmospheres would mean solving a cubic equation.
In a desire to make the function as realistic is possible, information for this problem has been taken from §5.8 of Zumdahl and Zumdahl, Chemistry, Houghton Mifflin, 5th edition, 2000. The first part of the expression for the pressure, $\frac{40}{2V-1}$, takes into account the inverse relationship between volume and pressure, but ignores the attraction between the gas molecules: this is reasonable as an approximation since, in its gas form, we expect lots of space between molecules and not much interaction. The molecules are, however, attracted to one another, and this attraction lowers the pressure. The term $-\left(\frac{4}{V}\right)^2$ accounts for this, providing a more accurate calculation of the pressure.
Instructors looking to increase the depth of the task might include a discussion about domain issues. It's clear that $V>0$ is necessary for the context (volumes are always positive), and also that $V=\frac{1}{2}$ is problematic for the function, and hence has to be excluded from the domain. Less immediate is that the interval $0<V<\frac{1}{2}$ has to be excluded, as those values of $V$ result in a negative value for the pressure, and that all values of $V>\frac{1}{2}$ do indeed give positive values of pressure. Without resorting to calculus (or an informal answer obtained by looking at the graph), the latter point can be argued by putting the expression for $P(V)$ over a common denominator and completing the square in the numerator.
Solution
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Below is a graph of $P(V)$:
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Notice that as the volume of the container increases, the pressure of the Xenon gas decreases which is what we would expect. As the volume decreases, the pressure goes up dramatically, with an asymptote at a volume of $\frac{1}{2}$ liter.
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Using the graph, the pressure appears to be 10 atmospheres when the volume of the container is close 2 liters, probably a little smaller. We can check how good the estimate is by substituting $1.9$ into $P$ and we find $P(1.9) \approx 9.9$.
Ideal Gas Law
A certain number of Xenon gas molecules are placed in a container at room temperature. If $V$ is the volume of the container and $P(V)$ is the pressure exerted on the container by the Xenon molecules, a model predicts that $$P(V) = \frac{40}{2V-1}-\left(\frac{4}{V}\right)^2$$ for all $V>\frac{1}{2}$. Here the units for volume are liters and the units for pressure are atmospheres.Â
- Sketch a graph of $P$.
- Using the graph, approximate the volume for which the pressure is 10 atmospheres.