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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant: $ PV = C. $

(a) Find the rate of change of volume with respect to pressure.

(b) A sample of gas is in a container at low pressure and is steadily compressed at constant temperature for 10 minutes. Is the volume decreasing more rapidly at the beginning or the end of the 10 minutes? Explain.

(c) Prove that the isothermal compressibility (see Example 5) is given by $ \beta = 1/P. $

a)$\frac{d V}{d P}=-\frac{C}{P^{2}}$$\\$b) at the beginning$\\$c) $\beta=\frac{1}{P}$

05:04

Smita P.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Sharieleen A.

October 26, 2020

Finally, the answer I needed, thanks Heather Z

Catherine A.

October 27, 2020

Thought I needed a tutor to help with Calculus: Early Transcendentals, but this helps a lot more.

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in this problem. We're using Boyle's law, and it has to do with pressure and volume pressure times. Volume is a constant when your temperature is a constant, and so we have PV equals C as the equation and for party, we want to find the rate of change of the volume with respect to the pressure. So first, let's rearrange our equation a little bit. Divide both sides by p, and we have volume equals the constant over the pressure. And we could rewrite that as volume equals a constant times of pressure to the negative one power that makes it easier to differentiate. So now we find the derivative of that. We bring down the negative one and multiply by the constant, and we raise P to the negative second power. Now we can rewrite that if we'd like to. As DVD P equals negative, see over p squared. In part B of this problem, the problem talks about how the sample is steadily compressed, and that means that the pressure is increasing. So if we take a look at our derivative that would just found in part A. If the pressure increases, that means you're having a larger denominator, and that's going to result in a smaller number. So what happens? As the pressure increases, the denominator grows, the fraction gets smaller, so DVD P the rate of change of the volume is heading toward zero. It's still negative, but it's heading toward zero. So that means that because the rate of change is getting closer to zero, the volume is decreasing more slowly as time goes on, so more rapidly in the beginning, more slowly at the end. And part C is about I so thermal compress ability The symbol for isil thermal compress ability is beta and an example five. In the book, it tells us that I so thermic compress ability is equal to negative one over v times DV 80 p What we just found in the first part that DVD P is equivalent to negative see over p squared so we can do that substitution. So now we have negative one over V times negative see over p squared. Let's simplify that. We can multiply the negatives together and we get positive, see over VP squared and we can call v p squared v P. Times p. And why would I want to do that because we know from the beginning of the problem that V, P or P V is equal to see the constant so it can substitute the constant in there. So now we have the constant over. The constant times of pressure canceled the constant and we have beta equals one over the pressure.

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