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A sample of xenon gas at a pressureof 1.12 atm and a temperatureof 28.7 °C, occupies a volumeof 14.3 liters. If the gasis compressed at constant temperature toa ...

Question

A sample of xenon gas at a pressureof 1.12 atm and a temperatureof 28.7 °C, occupies a volumeof 14.3 liters. If the gasis compressed at constant temperature toa volume of 10.6 liters, the pressure ofthe gas sample will be atm.

A sample of xenon gas at a pressure of 1.12 atm and a temperature of 28.7 °C, occupies a volume of 14.3 liters. If the gas is compressed at constant temperature to a volume of 10.6 liters, the pressure of the gas sample will be atm.



Answers

A sample of gas has an initial volume of 13.9 L at a pressure of 1.22 atm. If the sample is compressed to a volume of 10.3 L, what is its pressure?

I guess I mean to some questions too. Now we use Chancellor something special. Chapela relate for low pressure. Cass provided that picture is Constance So and the things they want, they want to Crete. Or I could make that. So the question have developed. They want to protect my mind later. They want to be while point to our most fair. Have me too to be 10.3 later on my feet to the floor. So from this formula, I can make my peace. Treaties are here. Maybe the one you want. Okay, if I eat too. So the next thing to do is just assorted too. The particles into tastes expression follow, Miss. We wanted starting from nine. The war on these one point to pack more spare Have we too? Is there a point? Three later. So people will be talking 0.95 point 25 Damn tree And I accuse while points his forces thiss Mitya Cancer Liver. That is the answer

This question. We have a sample of gas with an initial volume of 22.8 leaders and pressure of 1.65 tm. And then we're told the sample is compressed to a volume of 10.7. Leaders asked. Find the pressure. So we're going to use Boyle's law for this problem. Boyle's law is P one times anyone equals P two times B two, and we're looking for P two here. So p one I was 22.8 leaders. Also, that's the wonders. I also want to note that, um, leaders and atmospheres are the correct units per boils loss. We don't have to do any original, So if you want, it is 1.65 atmospheres. B one is 22.8 liters. P two was what we're looking for and you two is 10.7 leaders, so we do some math. Year 1.65 times 22.8 is 37.62 I'm a spear leader equal to P two times some 0.7 leaders. And if we divide by 10.7, we get P two is equal to 3.52 atmospheres

Hi there. This problem is located in the ideal gas equation section. So to solve this problem, we will be using the ideal gas law which is PV equals nrt. In this problem, we are given several values as we read through the problem. But we see that what we're trying to find, it's the number of moles. So we're trying to figure out how many moles of seen on gas we have reading on through the problem we see that were given a pressure of 1.18 atmospheres were given a temperature of 18 degrees Celsius. However, we know whenever we're using the ideal gas law equation, temperature must be absolute temperature. It must be in Calvin. So we're gonna add to 73 to this, which gives us 2 91 Kelvin and we're also given a volume of 26.7 leaders. So right now we have four out of the five variables in this equation we're missing. Our are is the universal gas constant. We have a couple of values for our that were given to us in this unit. We want to select the one that matches up in terms of units So since our pressure is an atmospheres, I'm going to select the R value that has atmospheres in its unit. So that is 0.8 to 1. Leaders times, atmosphere, leaders Sorry, times atmospheres over. Mole Kelvin has some rather odd units there, but we will see what happens with those units here. In a few moments, we are trying to solve for molds, which is end. So I'm gonna take the PV equals NRT equation before I put any numbers in it and I'm going to rearrange it. I'm going to isolate and by itself, by dividing both sides of the equation by rt If I divide both sides of the equation by rt I get end by itself and on the other side of the equation, I will get P V over R t. At this point, since I've identified all of our variables and our equation, we just need to plug these in and calculate the answer. So we have 1.18 atmospheres. We have a volume of 26.7. Leaders are we decided we were going to use the 0.0 8 to 1 leaders times atmospheres over mold Calvin and finally our temperature was 2 91. Calvin, have this all set up. It's just a matter of calculating it now. But before I do that, let's look at what happens to our units. Atmospheres will cancel. Leaders Will. Can't will cancel. Kelvin will cancel. The only unit we have left is moles, and it's in the denominator of the denominator. We have a complex fraction going on here, since it's in the denominator of the denominator. We know that that means it's actually in the numerator. So we go ahead and we calculate everything and we get 1.32 Moles of Xena. And that is our answer. Number of moles is 1.32 moles. All right, Thanks for watching.

In Problem 61 supports a 24.3 animal sample off helium gas at 25 degrees Celsius and 1.1 80 is heated to 50 degrees Celsius and compressed toe a volume all 15.2 ml. What is the pressure off? Simple. So this question is from the chapter guesses and the concept behind discussion is the combined Guess? No. So the combined gas law is the relationship between the pressure volume and temperature. So the relationship is even even divide by it even physical toe Pito way to divide by two. So this is the formula off the combined gas law or combined desecration and in this equation even represents the initial pressure off the guests. We even represents the initial volume off, I guess even represents the initial temperature off the guests. And similarly, the other terms P two Beato anti toe represents the final pressure off the girls final volume off guess and final temperature off the case respectively. So based on both formula, we can find out the final pressure. So the value off the one it's 1.18 p. M. The value off we one is 24.3 AM l and the value off the one is to 98. Basically it is 25 degrees Celsius and we convert into Calvin by adding to 73 which comes to 98. Tell me similarly, the value off Puerto we find out the value off rito given in the question 15.2 ml and the value off to do that is the temperature 50 degrees Celsius and we can avert in tow. Calvin So which is 3 23 Calvin. So put all the values in about formula, so we get 1.1 a tm into 24.3 Emmel divide by 2 98 Calvin Physical toe Pito whose value we find out volume is 15.2 ml divided by 3 23 Calvin. So on solving this, we find out the value of Butoh, which is near about 1.7488 p. M. Or approximately. We can write it at one point one point 748 p. M or 1.7558 a m. So this is the pressure off the sample so we can write so pressure off the simple is 1.748 p. M. So this is our final answer 1.748 p. M. And this is the but I shared off the sample.


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