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A $1-m^{3}$ volume of water is contained in a rigid container Estimate the change in the voleme of the water when a piston applies a pressure of 35 MPa....

Question

A $1-m^{3}$ volume of water is contained in a rigid container Estimate the change in the voleme of the water when a piston applies a pressure of 35 MPa.

A $1-m^{3}$ volume of water is contained in a rigid container Estimate the change in the voleme of the water when a piston applies a pressure of 35 MPa.



Answers

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 m$^3$ of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 m$^3$. If the temperature remains constant, what is the final value of the pressure?

Original volume of the air inside the cylinder is 0.110 cubic meter Evaporation off 3.40 atmosphere. So here we have the initial physical toe Jura pie in 110 giving meter on DDE be initial physical too. 3.4 Tzeitel atmosphere Now the piston ob cylinder is pulled out so the new volume off air becomes zero pirate 390 cubic meter. If the expansion happens at Constant on Bridger, then we are asked to calculate the new pressure inside the cylinder. According to Arial gay seclusion, we have BB physical toe and times our times. Do you, in this case number of moves, Uh, and a doozy. Costin So vo BB physical to constant Are we can I did Yes, be final times the final physical do be initial times the initial From here we have far be of is it will do b i times we eye upon We have no but liking different values We have we have physical do here be eyes 3.40 atmosphere multiply re eyes 0.110 Ridicule upon Rieff He's zoo defiant 39 zero You weak meter So we good be a physical toe zero pliant 96 atmosphere So their sensor Thanks for watching

Now here we have to find a doorbell oppression which he's atmospheric pressure and plus the extra patient. The exhibition is force over area fourth is mg and the area is by our sled. It was atmospheric pressure is 14013 I am standing five baskets. Thus mass. Want to indicate a brand jeans I bought a by the radio is 6.1 centimeter all pseudo 0.61 major split off that. So from here we find that be daughter equals you tend to the five fastball So from the vaporization card we have the desire temperature to be from 20.

For this problem will set the density of water at 10 degrees to know or not, And that density is 999. What 73 Katie or cubic? Mita. The density then, and 100 degrees is 900 and 58. 138 80 per cubic meter. And you call this value rope. So from here, follows at the volume of water, and 100 degrees C is equal to a massive water em over its density. Rho and the north. 10 degrees is equal to the same mess. Remember, the mess doesn't change, but the density does, since the volume does so that's him were or not. So what this tells us is that doubts of E, which is the changing for you at Tunis and temperatures is able to em in two B minus V note, which is one over no minus one over the noise. We also know that's a volume expansion and be expressed. That, to me, is equal to be not Beatle repeater is the coefficient off volumetric expansion. I'm not a tea without tea with the change in temperature. So, in this problem we reached assault for beat up when you're quite efficient of suspension. And so if you're hearing this creation, we can see that be talk is equal to I m over. Be not don t into one or row minus one over or not. But m over not is simply not our density of water tech leads. That's right. Not over a change in temperature out t into one over no minus y over or not Now all of these values that no So hence we can count calculated value beat up in this temperature range. So no, not is 900 in 99 0.73 You could buy this by temperature change off 90 Southeast agrees on 19 Taliban suppressed the units here into one over the final pressure throw off 958 0.38 minus one over north, which is 999 0.70. And so completing this we get the average coefficient of volume expansion Peter in this temporary temperature range useful 0.79 I'm stand to the minus full Calvin or a Southwest degree

Hi there. So for this problem we need to calculate the change in death that causes the pissed um to move better, an amount of 0.70 centimetres. So the information given are the force constant or the springs constant? Were also given the the diameter of the piston. We are also given the radius of the piston and that the information of the length or yes, the distance that the spring has moved. So the first time to consider for this problem is that when the gauge is lower, the change in pressure, we know that from the definition of pressure is given by the four supply over the area. So the only force that is being applied in here is the force due to the spring. And we know that that is the sprints. Constance times the distance it has moved and the area of application, which is p times are square because that's the um barrier of them piston. And we can obtain this. Since we have all of this information, We know that constant is this value. We know that it has made a value of ads with 0.75 time standard and minus two m. Remember that you need to put those in the current units that we can eliminate that. And in the denominator we have pi times the radius square from the picture. We know that this radio is still 0.6 times 10 to the minus two um meters in these squared. So uh from here, when we pull out of this into the calculator, we obtain a value for the change in pressure off 8.28 times 10 to the fourth. Newtons per meter a square. So this is the change in pressure. Now we know that we are um under under L. M. We are submerged in water in this case. So that will know that there is a pressure because of the liquid. Um That is given by the density of the liquid. The acceleration due to gravity and the hate or the death the system it's submerged. So the density of water we know that is one approximately 1000 kg per meter Q. And we know the acceleration due to gravity on earth is 9.8 m per second square. And we know and we need to solve for age. That's the one that we need to determine for this problem. So that will be the change in pressure over the density times the acceleration due to gravity. So we have obtained that the pressure or the change of pressure is eight point 28 and the density is one person and the gravity. So with this plug this into the calculator and we obtain a death of eight point five m. So this is the death and that the system must be um submerging water so that the peace doesn't move by an amount of 0.70 centimeters. So this is it for this problem. Thank you.


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