# Boyles Law, Graphs

by TheRedDevil18
Tags: boyles, graphs
 P: 168 Hi, what would be the relationship between the product of pressure and volume (pv), against pressure. How would you represent it on a graph?, I have heard that the shape will be a horizontal line. Could someone please show an example or explain it to me?
 P: 485 See, for example, this website at NASA. You should feel free to plot the product of pressure and volume against pressure given their demonstration for yourself. You could also plot the product against volume for completeness's sake.
 P: 168 Thanks, with the graph volume vs 1/pressure, I should just take the pressure and invert it and then sketch the graph eg. pressure 150kpa = 1/150 = 0.007. Is their any other calculations?
P: 854
Boyles Law, Graphs

 Quote by TheRedDevil18 Hi, what would be the relationship between the product of pressure and volume (pv), against pressure. How would you represent it on a graph?, I have heard that the shape will be a horizontal line. Could someone please show an example or explain it to me?
The product PV for an ideal gas is constant, assuming no excess moles were added. So if you plot it against the pressure or the volume, it gives you a horizontal straight line.

 Thanks, with the graph volume vs 1/pressure, I should just take the pressure and invert it and then sketch the graph eg. pressure 150kpa = 1/150 = 0.007. Is their any other calculations?
Yes, just calculate the corresponding 1/P for V. The graph is of the equation,

$$V = \frac{k}{P}$$

$$PV = k$$

From mathematics, this comes out to be a rectangular hyperbola.
 P: 168 So in my example pressure = 150 kPa and Constant = 1530, it will become 1530/150 = 10.2 which is my volume, so its like plotting volume against volume, so my graph is a straight line?, Are you sure this is correct? and also what does the gradient represent, is it temperature?
P: 854
 Quote by TheRedDevil18 So in my example pressure = 150 kPa and Constant = 1530, it will become 1530/150 = 10.2 which is my volume, so its like plotting volume against volume, so my graph is a straight line?, Are you sure this is correct? and also what does the gradient represent, is it temperature?
Whoops. You do calculate the 1/P, for V and plot the graph. I misinterpreted your previous post and told you the hyperbolic relation for P and V, and not 1/P and V..

The graph between 1/P and V would simply be a straight line, as it is of the form y=mx ( or V=k/P). This line passes through the origin and has the slope k.
 P: 168 So to get 1/P, it would be V=k/P, right?, and does the gradient represent temperature?
P: 854
 Quote by TheRedDevil18 So to get 1/P, it would be V=k/P, right?, and does the gradient represent temperature?
Yep. The gradient does not represent temperature, though. Since gas is ideal and assuming one mol,

$$k = PV = RT$$

So, k is proportional to temperature, though not exactly equal to temperature.
 P: 168 So k = RT, what does RT represent?
P: 854
 Quote by TheRedDevil18 So k = RT, what does RT represent?
Do you know the Ideal gas equation?

R is the universal gas constant, and T is temperature of the gas.

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