A 20.0-cm-long, 10.0-cm-diameter cylinder with a piston at one end contains 1.34 kg of an unknown liquid. Using the piston to compress the length of the liquid by 1.00 mm increases the pressure by 41.0 atm. What is the speed of sound in the liquid?

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Hey everyone in this problem, a rectangular tank measures 1.5 m by 1.2 m by m. The cross section 1.5 m by 1.2 m has a lid thick and compress liquids in that tank. We're told that moving the lid by 0.25 m to compress, compress a 90.6 kg liquid in the tank raises the liquids pressure by 1.5 times 10 to the exponent six plus gas. We asked to calculate the speed of sound in this liquid. We have four answer choices all in meters per second. Option A 691. Option B 319 option C 40. and option D 214 we're looking for the speed of sound. We're giving some information about a change in pressure, a change in volume. So let's recall two equations. First tells us that the change in pressure, delta P is equal to the bulk modulus B multiplied by the change in volume, delta V divided by the volume. We're gonna call this equation one, we have the second equation. And the second equation is that the speed of sound a little V. Let me draw this as and I'm gonna call this vs just so that it's clear that this is the speed of sound and not the volume. OK? Because we have both of those vs in our equation. So we're gonna call this vs is equal to the square root of the bulk modulus B divided by the density ro All right. So this is gonna be equation now, why have I written both of these equations? Well, we're getting information about the change in pressure, the change in volume and the volume, but we don't know what this liquid is. So we don't know the bulk modules. We can't compute the bulk modules. However, we're looking for the speed of sound. So what we can do is use equation two to write the bulk modules B that we don't know in terms of the speed of sound, substitute that into equation one. And then we'll have an equation where we can solve for that speed of sound BS. And we've gotten rid of the bulk modules B that we don't know. All right. So from equation two, we can write the bulk module is A B is going to be equal to V squared multiplied by and we squared both sides and then multiplied by a broke. And this is vs squared. All right. So let's substitute this into equation one. We can write that delta P is equal to the bulk modulus which is now vs squared right multiplied by delta V divided by B. We want to solve for the speed of sound of es we can multiply both sides by the volume B then divided by row delta B, we get that vs square speed of sound square is equal to delta P. The change of pressure multiplied by V, the volume divided by row density multiplied by delta V. The change of volume. The final step is gonna be to take the square root to completely isolate for that speed of sound. So vs is gonna be the square root of this entire expression. Delta P multiplied by V divided by a row delta P. All right. So I said there'll be no information about all of these things. So let's go ahead and write them down. OK. Now delta P is the change in Russia. We are told in the problem that the pressure changes by 1. times 10 to the exponent six pascal. So delta P, this known value is gonna be equal to 1.5 times 10 to the exponent six. What about the volume? The volume? This is gonna be equal to the initial volume of our tank. OK. So we have a 1.5 m by 1.2 m by 2 m. So we have a length multiplied by width multiplied by height because we have a rectangular type. This is gonna be equal to 1.5 m. Multiplied by 1.2 m multiplied by m which is equal to 3. meters. Cute. The next thing we wanna find is delta be the change in volume. We're talking about the change in volume. Ok? We're told that we move the lid by 0.25 m. So we go from a 1.5 m by 1.2 m by 2 m tank to 1.5 by 1.2 by two plus or minus 0.25. Ok. So the change in volume, delta B, it's just going to be equal to now the length multiplied by the width multiply by the change in height and the length and width remain the same, the height changes. So the change in volume is due solely to that change in height. This is gonna be equal to 1.5 m multiplied by 1.2 m multiplied by 0. m. And this gives us a change of volume of 0. m. All right. So delta PV and Delta V are done. Now, we need to find the density. Now we aren't given the density, but we are given information about the mass. We know the volume. So let's recall it, the density row gonna be equal to the mass divided by the volume in this problem. That is 90.6 kg divided by the volume. We found of 3.6 meters. Cute, you get a density of 25. repeated kilograms per meter. All right. So now we have all of these values that we needed to substitute into our equation for VS. So let me move. So we can see all of them at once. We're gonna get back to our vs equation and substitute in all of these values. So vs is gonna be the square root of 1. times 10 to the exponent six multiplied by 3.6 m cubed divided by 25.166 repeated kilogram per meter. Cue 550.45 meters. Cute. All right. Now, when we talk about units in the denominator, these units of meters cubed, we're going to divide it. Recall that a pascal is equivalent to a kilogram per meter second squared. OK. So we have a kilogram per meter second squared multiply by meters cubed and then divide by a kilo. So the unit of kilogram will divide up one of those meter terms will divide it. And we're gonna be left with meters squared per second squared inside of our square root. And it's always good to check your units to make sure that they match up with what you want. All right, so we have it with meters squared per second squared inside the square root. When we take the square root, you get meters per second, which is what we want for the speed. And we find that the speed of sound here is gonna be approximately equal to 690. m per second. If we compare this with our answer choices around to the nearest meter per second and our answer corresponds with option a 691 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

A 20.0-cm-long, 10.0-cm-diameter cylinder with a piston at one end contains 1.34 kg of an unknown liquid. Using the piston to compress the length of the liquid by 1.00 mm increases the pressure by 41.0 atm. What is the speed of sound in the liquid?

Verified Solution

Key Concepts

Pressure and Volume Relationship

Speed of Sound in Fluids

Bulk Modulus