(II) How much pressure is needed to compress the volume of an ...

Question

(II) How much pressure is needed to compress the volume of an iron block by 0.10%? Express your answer in N/m², and compare it to atmospheric pressure (1.0 x 10⁵ N/m²).

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the bulk modulus in this problem, an iron bar needs to be compressed by 0.2% by pressure. And the question wants us to compare the atmospheric pressure which is approximately two multiplied by 10 to the five newtons per meter squared. With we with the answer that we calculate for the pressure in newtons per meter squared. We're also given a hint that tells us that the bulk modulus of iron is 1.6 multiplied by 10 to the 11 noons per meter squared. We're given four possible choices and each answer choice has two parts to it. So for choice, A we have 3.2 multiplied by 10 to the eight moons per meter squared. And the second part is 1.6 multiplied by 10 to the three atmospheres. B is 4.3 multiplied by 10 to the eight means per meter squared. And the second part is 2.3 multiplied by 2 to 3 atmospheres C is 5.4 multiplied by 10 to the eight noons per meter squared. The second part is 3.4 multiplied by 10 to the three atmospheres. And choice D is 6.7 multiplied by 10 to the three noons per meter squared. And the second part is 5.6 multiplied by 10 to the three atmospheres. The first thing we're gonna do is recall our definition for the bulk modules since we were given that in the problem. So the bulk modulus B is equal to minus it change of pressure, delta P divided by the quantity of delta V over V knot. So delta V is our change in volume. And V knot is our original volume. We're interested in calculating the pressure. So we'll solve this for delta P. In that case, I'll have delta P will be equal to minus B multiplied by the quantity of delta V over divided by VN. Now we were given the bulk modules and the problem, but we don't have any information to calculate our volume. However, we are told that the iron bar is being compressed and we're given the percentage by which it's compressed, which is 0.2%. So we're gonna write our delta V in terms of this percentage we'll have for our delta V that's gonna be equal to. And we're told that it's compressed, that means my final volume is smaller than my initial volume. So when I look at my delta B here, that's gonna be final minus initial, that's gonna be a negative quantity. So I have delta V is equal to minus. And here I'm going to have this change be a percentage of my original volume. Not. And I'm told that percentage and that is um the 0.2%. So I'll change that to a decimal. That's gonna be 0.002 multiplying the not so to change the percentage to a decimal just divided by 100. So I can plug that back into our formula here for our pressure, they will have delta P is equal to minus e multiplying the quantity of negative 0.002 VK NOTT all divided by V knot. To hear my original volume cancels out the V knots cancel, which is great because we didn't have any information to calculate that quantity. And what I'm left with is our delta P is equal to 0.002 multiplied by B. So I was given value for the bulk modulus. So I'll plug that in the delta P is equal to 0.002 multiplied by the quantity of 1.6 multiplied by 10 of the 11 newtons per meter squared. And so when I calculate my change in pressure, which is gonna be how much pressure needs to be applied to get this um change in volume. This turns out to be 3.2 multiplied by 10 to the eight students per meter squared. Now, the second part to this question wants us to compare this to the atmospheric pressure. So what I want to do is calculate the ratio of my delta P divided by the P of the atmosphere. And so I have values for both of those. So the delta P that I just calculated that's gonna be the 3.2 multiplied by 10 to the eight newtons per meter squared. And I'll divide that by the, the two multiplied by 10 to the fifth Newton's per meter squared. And so when I do that, I get the delta P. Bye, bye. The atmosphere. That's gonna be equal to 1.6 times 10 to the three. And so what that means is my delta P is gonna be 1.6 multiplied by 10 to the three atmospheres. And so those are the two parts of the question that I'm looking for. And if I look at the choices that corresponds to answer a, so just to recap what we've done real quick, we used our definition of the bulk modulus to find our change in pressure. But in order to do that, we needed to first um find our change in volume in terms of our original volume. And we also had to take into account the fact that this was being compressed. So that means our change in volume is actually going to be negative. So I hope that this has been useful and I'll see you in the next video.

(II) How much pressure is needed to compress the volume of an iron block by 0.10%? Express your answer in N/m², and compare it to atmospheric pressure (1.0 x 10⁵ N/m²).

Key Concepts

Pressure

Bulk Modulus

Comparison to Atmospheric Pressure

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(II) How much pressure is needed to compress the volume of an iron block by 0.10%? Express your answer in N/m2, and compare it to atmospheric pressure (1.0 x 105 N/m2).

Key Concepts

Pressure

Bulk Modulus

Comparison to Atmospheric Pressure

Watch next

(II) How much pressure is needed to compress the volume of an iron block by 0.10%? Express your answer in N/m², and compare it to atmospheric pressure (1.0 x 10⁵ N/m²).