5

Assume that you have a sample of gas at 350 K in a sealed container; as represented in part (a): Which of the drawings (b)-(d) represents the gas after the temperat...

Question

Assume that you have a sample of gas at 350 K in a sealed container; as represented in part (a): Which of the drawings (b)-(d) represents the gas after the temperature is lowered from 350 K to 150 K if the gas has a boiling point of 200 K? Which drawing represents the gas at 150 K if the gas has a boiling point of 100 K?4b)(di

Assume that you have a sample of gas at 350 K in a sealed container; as represented in part (a): Which of the drawings (b)-(d) represents the gas after the temperature is lowered from 350 K to 150 K if the gas has a boiling point of 200 K? Which drawing represents the gas at 150 K if the gas has a boiling point of 100 K? 4b) (di



Answers

Assume that you have a sample of gas at 350 $\mathrm{K}$ in a sealed container, as represented in (a). Which of the drawings (b) - (d) represents the gas after the temperature is lowered from 350 $\mathrm{K}$ to 150 $\mathrm{K} ?$

Hi there in problem number two. We have some gas in a movable piston. So this part right here is movable. It can slide up or down and our gas is here in the piston. So, we have three different scenarios here that they would like us to describe what is going to happen in each of these scenarios. And we need to redraw this for each of these scenarios. So let her a let me draw the container. Mhm. Okay. In letter A. It says the pressure is remaining constant, But the temperature of the gas is being increased from 300 to 500 Kelvin. So when temperature goes up and pressure remains constant, the volume is directly proportional. It will increase By the same proportion. So going from 300 to 500 is not quite doubling. So, if our original piston was here, our new piston height is probably right about up here. Make that center longer. All right. And we have our gas particles in there because they have heated up. They are moving faster. They have more kinetic energy and they will cause an expansion in the container. All right, moving on to letter B, drawing our container. Now the temperature remains constant and be but the external pressure on the piste and has increased From 101.32 to 2.7 while temperatures constant. So in other words, the pressure, the outside atmospheric pressure has approximately doubled, so pressure and volume of a gas sample are inversely proportional. If pressure increases and by doubling the pressure or the volume is going to be cut in half. So that means the piston will get pushed further down. I just cut the original volume in half. Or tried to anyway. And we still have our gas particles in here, Right? And then finally let her see same gas sample in that same container. But now The temperature is decreasing from 300 to 200 while the pressure remains constant. So temperatures decreasing. The original piston was right about here, Going from 300 to 200, so it's going down by a third. So if the temperature decreases by 1/3, The volume will decrease by 1/3. So that was my attempt to show the volume of the container At 2/3 what it was originally. So this is approximately what these pistons will look like after each of these scenarios.

So for part A we have a container with a piston And we're at one atmosphere in 300 Kelvin and we're going to change the volume because they're going to increase the temperature. Okay well volume and temperature are directly proportional. So you could say the one over T one equals B to over T two. So be one. So the volume at 300 Kelvin, Let's compare it to the volume at 500 Kelvin. So the new volume Should be 5/3 of the original volume. So let's draw our piston higher so our volume is going to be that much bigger but we still have the same number of gas particles in here. Okay, but that will be our new volume and then we're going to change our pressure, we're going to double it. Okay, you're gonna increase our pressure so pressure and volume are inversely related. So we could write P one V one equals P two V two, So one times v one Because the pressure of the two. So our new volume Should be 1/2 our original volume. So we were out there so we'll go about here now that's about half the volume. Okay, so there's our piston but we still have the same number of particles in here and that's why there's a larger pressure

This question shows a picture of a gas in a container with a piston, something similar to what's shown here. It then wants you to buy drawing pictures demonstrate your conceptual understanding of what happens if the temperature changes and nothing else, or if the pressure changes and nothing else. If the temperature is going to increase from 300 to 500 Kelvin, almost doubling To the point where it's 5/3 the original temperature. Then in order to keep the pressure constant, the volume must increase to 5/3 the original volume. If for part B, the pressure is doubled while the temperature stays constant. The only way this can occur is if the volume is cut in half And for the last one, if the temperature is decreased as temperature decreases, assuming pressure stays constant, then volume will also decrease. So if the value of the temperature goes from 300 to 200 and we are now at 2/3 the temperature, so 2/3 the volume.

Okay, So in this question, we have a piston and it's undergoing certain changes. And, ah, if they're asking us to redraw what the pistol would look like after these changes have taken effect. So in the 1st 1 we can see that we're going from the temperature of 300 Kelvin to a temperature of 450 Kelvin. So Ah, this is already in Kelvin. It's an absolute temperature, which means that we can treat it proportionally. So if we look at this, the temperature is increasing by 50% which means that the volume would also increase by 50%. So we'd be right around here in the 2nd 1 The pressure is increasing from 1 to 2. So essentially, what we're saying if we're pushing the piston down further to increase that pressure to increase the force on that gas and ah, if it if the pressure is doubled, that means the volume is going to be cut in half. So this is gonna be right about at the halfway point. So it's has gone down 50 percent. 1st 1 went up 50%. This one went down 50%. And then finally, we have the temperature is decreasing by about 1/3 by exactly 1/3. And the pressure is also decreasing by exactly 1/3. So the temperature decreasing by 1/3 is going to shrink our volume. But the pressure decreasing by 1/3 is going to grow our volumes. So in this case, these two are actually gonna cancel each other out. So we should end up with our piston exactly where it started because we're multiplying by 2/3 and then we're dividing by 2/3 so it's gonna cancel out.


Similar Solved Questions

5 answers
Calculate the height in meters of a column of water that exerts a pressure of 2.00 bar at its base Express your answer using three significant figures:Azdh =m
Calculate the height in meters of a column of water that exerts a pressure of 2.00 bar at its base Express your answer using three significant figures: Azd h = m...
5 answers
2-SampTInt.LQQUD n= |00 XI-bar 128 S1 = 12ETQuD 2 m = |05 X-bar = 132 52 =20The point estimale X, - Xz95% confidence interval (rounddecimal places):margin of eitorIs the diflerence m nkans signifiant? Why or why not?
2-SampTInt. LQQUD n= |00 XI-bar 128 S1 = 12 ETQuD 2 m = |05 X-bar = 132 52 =20 The point estimale X, - Xz 95% confidence interval (round decimal places): margin of eitor Is the diflerence m nkans signifiant? Why or why not?...
4 answers
Iflx) = (x-aJr-bJx-c), show that f (r) = 1_+3 f(x) X-a x-b r-c
Iflx) = (x-aJr-bJx-c), show that f (r) = 1_+3 f(x) X-a x-b r-c...
5 answers
Question 126.25 ptsThe decomposition of NO and NOz to the elements proceeds via the following two reactions: 2NOlg) + Nzls) Oz(g) AHrxn -169 kJ 2NOzlg) 7Nzlg) 2O2(g)AH,xn 46 kJ Calculate the enthalpy change in kJ for the reaction below using these information:2NOzlg) + 2NOlg) Ozlg)
Question 12 6.25 pts The decomposition of NO and NOz to the elements proceeds via the following two reactions: 2NOlg) + Nzls) Oz(g) AHrxn -169 kJ 2NOzlg) 7Nzlg) 2O2(g)AH,xn 46 kJ Calculate the enthalpy change in kJ for the reaction below using these information: 2NOzlg) + 2NOlg) Ozlg)...
5 answers
Diagram; label the location of the frce cnergy asscciated with the; substate [- In the following enzyme-substrate complex activation energy wlo enzyme product
diagram; label the location of the frce cnergy asscciated with the; substate [- In the following enzyme-substrate complex activation energy wlo enzyme product...
5 answers
Which one of these is a best fit for a Kaizen project (Lean)?a. Design of a new laboratoryb. Improving the flow of a laboratoryc. Improving the turnaround times for the laboratoryd. Reducing laboratory billing errorse. Both $mathrm{b}$ and $mathrm{C}$
Which one of these is a best fit for a Kaizen project (Lean)? a. Design of a new laboratory b. Improving the flow of a laboratory c. Improving the turnaround times for the laboratory d. Reducing laboratory billing errors e. Both $mathrm{b}$ and $mathrm{C}$...
5 answers
IW WT 0 J0 ajueisip pOOM Jo xjojq e sajennauad 1! se s/u 087 we BullaneJ} Jalinq 84 ST"0 e do1s 01 Auessajau S! 33J0y a8eua1e JeyM
iW WT 0 J0 ajueisip pOOM Jo xjojq e sajennauad 1! se s/u 087 we BullaneJ} Jalinq 84 ST"0 e do1s 01 Auessajau S! 33J0y a8eua1e JeyM...
5 answers
4. The body-mass index (BMI) of all American 20-something women is believed to follow a Normal distribution with a standard deviation of 6.2. Find how many 20-something women should be randomly sampled if we want the mean BMI of such a sample to be within BMI unit of the population mean with 95 percent confidence_ (15)
4. The body-mass index (BMI) of all American 20-something women is believed to follow a Normal distribution with a standard deviation of 6.2. Find how many 20-something women should be randomly sampled if we want the mean BMI of such a sample to be within BMI unit of the population mean with 95 perc...
5 answers
Suppose that a particular company estimates its growth income bythe formula: dS/dt=2(t−1)2/3 where S millions ofdollars in the growth income from sales t years hence. If thegrowth income from the current year’s sales is 8$ million, whatshould be the expected gross income from sales two years fromnow?
Suppose that a particular company estimates its growth income by the formula: dS/dt=2(t−1)2/3 where S millions of dollars in the growth income from sales t years hence. If the growth income from the current year’s sales is 8$ million, what should be the expected gross income from sales ...
5 answers
(1 point) Does there exist a continuous function f (x) such that f (0) ~9_ f(2) = -3 and f' (x) < 7 for all € in (0,2 )2Answer: yes or no Note: You only have one chance to input your answer:
(1 point) Does there exist a continuous function f (x) such that f (0) ~9_ f(2) = -3 and f' (x) < 7 for all € in (0,2 )2 Answer: yes or no Note: You only have one chance to input your answer:...
5 answers
(6) Find the conditional probability mass function Pxix (2/2)-
(6) Find the conditional probability mass function Pxix (2/2)-...
5 answers
Achareed particle Q Is moving with velocity 1 In & circle with radlus R under the Influence 0f magnetlc fleld B. The work done on Ihe particle In every cycle is:Select one:ZTRQvB RQvBTRQvBZero
Achareed particle Q Is moving with velocity 1 In & circle with radlus R under the Influence 0f magnetlc fleld B. The work done on Ihe particle In every cycle is: Select one: ZTRQvB RQvB TRQvB Zero...
5 answers
Find an equation of the tangent line to f(x) = x/ x^ 2 + 1at x = 1.
Find an equation of the tangent line to f(x) = x/ x^ 2 + 1 at x = 1....

-- 0.029311--