At constant T and n, the pressure and volume of an ideal gas are inversely proportional to each other. A graph of V vs P is hyperbolic, while a graph of V vs 1/P is linear. Make sketches of these relationships, then find an expression for the slope of V vs. 1/P by rearranging the ideal gas equation to the form V = slope·(1/P)
Please explain well! Thank you!
The ideal gas equation is written as
Where
P = pressure
V = Volume
n = Number of moles of gas
R = Gas constant
T = temperature in Kelvin
When number of moles of gas (n) and temperature T are constant, the right side of the equation becomes and constant as R is already a constant. Also, the product of three constant is also a constant.
Hence, we can write
Where the constant is nRT.
Let's take a simple real world example to visualize this. Assume there is a cylinder fitted with a movable piston filled with Argon gas. If we move the piston downward, the volume of the cylinder will decrease. As the temperature and number of moles of the Argon gas is not changing, the molecules will now collide with each other and the wall of the cylinder frequently as the volume has decreased, which will lead to an increase in pressure. Also, when the piston is moved upward, the volume of the cylinder will increase, hence, the area of the cylinder wall also increases, it means there will be less pressure as force per unit area of the cylinder will be lowered. Hence, pressure and volume are inversely related given that the number of moles of gas and the temperature are constant.
Now, if we plot V vs P by taking V in y axis and P in x axis, we get the following graph:
Note that as value of P increases, the volume V decreases.
Now, we can plot V vs 1/P.
Hence, in the above plot, the equation of the straight line is
But since the line passes through origin, the intercept c= 0.
From the ideal gas equation above, we know that the slope of the graph will be nRT.
Hence, we can reliably get back the ideal gas equation by putting in slope = nRT which proves the relation.
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