A player bounces a basketball on the floor, compressing it to 80.0% of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at 20.0°C and 2.00 atm. The ball's inside diameter is 23.9 cm. (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal.

Video transcript

Hey everyone in this problem. We have compressed air and a container fitted with a piston K. It's used as a shock absorber in one instant, the air is compressed to 70, of its initial volume. As it absorbs shock. The shock absorber initially contained inert and two gas at a pressure of 4.5 atmospheres and 25°C. Okay, the container is a cylinder with a diameter of .18 m, a height of .42 m. And we are asked to find the temperature of end to at the compression. Okay, this is an automatic process. An end to is considered ideal. Okay. All right, so audio batic process, we think of no heat transfer. Okay, so Q is zero. Now we're looking for temperature when we have an automatic process, let's recall that we have an equation dealing with temperature that tells us T. One V. One to the gamma minus one is equal to T two. V 2 to the game of -1. Okay, Alright. We have T. One. Okay, so T. One, we'll just right over here we're giving us 25 degrees Celsius, Which is going to be equal to 298.15 Calvin. We don't know V one but we do have a relationship between V one and V two and we're looking for that temperature T. Two. Okay, so we're looking for T. Two. So, let's go ahead and fill in the information. We know about the relationship between V one and V two and see where that gets us. Okay, so on the left hand side we're just gonna have T. One V. One to the gamma minus one. Like before on the right hand side we have T. Two and then we have V. Two which we know is 75% of the initial volume. Okay, so this is going to be 0.75 times V. One all to the gamble minus one. All right. We want to solve for T. two. So let's go ahead and isolate T. two. So dividing we get T. Two is equal to T. One V. One to the game of minus one all divided by 0.75. V. One to the gamma minus one. Now when we have things multiplied together to an exponent we can split those. So we have T. One V. One to the gamma minus 1/ 10.75. To the gamma minus one times V. One to the gamma minus one. Ok. And you'll notice here that the V. One gamma minus one will divide out. Okay, this leaves us with T. Two is equal to T. 1/0 20.75. The gamble minus one. Okay, so it turns out we don't even need to know the exact volume initially. Okay, we just needed the relationship between the two. Now we know T one. The only thing left to find is gamma. Okay, once we know gamma will be able to substitute to find T. Two. Now let's recall that we have a relationship gamma between C. P. And C. V. Mhm. Now we are told that we have n. two gas. Okay the two tells us that this is diet atomic. Okay if this is di atomic that means that C. V. Is going to equal five halves times. Are okay so we're looking for gamma. We know C. V. Now what about C. P. Well this is ideal gas. Okay we're told that in the problem so we can recall that CP is going to equal Cv plus our. Okay well this is just gonna be five halves R plus R. Which is gonna be seven halves are so now we have C. V. We have C. P. So we can go ahead and find gamma. Gamma is going to be the ratio of the tube. So it's gonna be seven halves are Divided by 5/ are Which is going to be 7/5. Okay so now we have our temperature T. One. We have our gamma value 7/5. We can substitute them in to find T. Two. And we're gonna get T two is equal to 298.15 Calvin Divided by 0.75 To the 7/5 -1. Okay this is going to give us an exponent of 0.4 and this is going to give us T2 is equal to 334.51 Calvin or equivalently 61.36°C. Okay, So this is the final temperature that we were looking for. Okay. And that's going to correspond with answer. C 61°C. Thanks everyone for watching. I hope this video helped see you in the next one.