Boyle's Law Question/Problem

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Hi Everyone, first semester Physics student here and we are working on a tough lab this week, having to do with Boyle's Law...argghh!!! :cry:

Anyways, here is my problem, and trying to work through it to obtain the correct answer...but it is stumping me!!

The question as follows:

1. Suppose data was collected for a sample of 3.oo moles of the ideal gas represented by the Pressure versus Volume -1 graph below. What is the Kelvin temperature of the gas?

the graph has P (pressure) for the Y axis, and 1/V for the x axis.

The slope =45.6 atm L and is at an approximate 45 degree angle (straight line) on the graph.

Any thoughts?

I am trying to use the equation of P=nRT/V=constant/V to solve this, but am not sure where/how to start.
P=total pressure of the gas
V=Volume of the gas
n=# of moles of gas
T absolute temperature og gas in Kelvin

It is given that R=8.314 J/mol and K=0.0821 atm liter/mol K

And here is a real tough one...

2. 2.00 moles of an ideal gas at -50 degrees Celsius has a volum of 1.00 liter. What is the pressure (in atm) in the container??

I can't find the proper equation/setup for this one and am totaly lost here...these two problems are just evil!!!

Please help, if you can!! Thanks!! :smile:
 

Answers and Replies

  • #2
Hootenanny
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Well, you can use your equation;

[tex]P = \frac{nRT}{V}[/tex]

Now consider the general equation of a straight line;

[tex]y =mx +c[/tex]

Think about what the gradient of your line represents, it may be more useful if we re-write your equation;

[tex]P = nRT \cdot \frac{1}{V}[/tex]

Regards,
-Hoot

[Edit] I've just seen your other post now thunder, welcome to PF![/Edit]
 
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  • #3
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Hi Hoot, for some reason, when I click on this topic, I can't pull up your reply. just joined today so I may not be hitting the right buttons yet; Cool I see it now!! Hey thanks, hoot!! Great to be here!! :)
 
  • #4
Hootenanny
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For your second question, you need only apply the equation

[tex]pV = nRT[/tex]

Ensure that the volume is in [itex]m^3[/itex] and the temperature is in kelvin. Also note that 1atm is approximately [itex]1.01\times 10^5[/itex] pascals.

Reagards,
-Hoot
 
  • #5
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Hootenanny said:
Well, you can use your equation;

[tex]P = \frac{nRT}{V}[/tex]

Now consider the general equation of a straight line;

[tex]y =mx +c[/tex]

Think about what the gradient of your line represents, it may be more useful if we re-write your equation;

[tex]P = nRT \cdot \frac{1}{V}[/tex]

Regards,
-Hoot

[Edit] I've just seen your other post now thunder, welcome to PF![/Edit]
Ok so here is my calculation method for question #1:

Suppose data was collected for a sample of 3.00 moles of the ideal gas represented by the Pressure versus Volume -1 graph below. What is the Kelvin temperature of the gas?

the graph has P (pressure) for the Y axis, and 1/V for the x axis.

The given slope = 45.6 atm L and is at an approximate 45 degree angle (straight line) on the graph.

Using equation P=nRT x 1/V

slope = 45.6 atm = nRT

R=universal gas constant=8.3145 J/mol K

n=# of moles=3.00 moles for this problem

so, I solve for T as follows:

nRT = 45.6

(3.00)(8.3145)T=45.6

DIVIDE both sides by (3.00)(8.3145) OR (24.9435)

T= 45.6/(24.9345)

So then solving for T=1.82813157736 degrees Celsius

So my answer in Kelvin would be 1.82813157736 + 273

Which is approximately 274.8 Kelvin

Is this correct? The question is asking for me to obtain the correct Kelvin temperature of the gas.
 
  • #6
Hootenanny
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Not quite. As the gradient is given in atm L / mol, you need to use that form of the universal gas constant (i.e 0.0821). Also the T in the equation represents absolute temperature, so there's no need to add 273 to your answer.

Regards,
-Hoot
 
  • #7
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Hootenanny said:
For your second question, you need only apply the equation

[tex]pV = nRT[/tex]

Ensure that the volume is in [itex]m^3[/itex] and the temperature is in kelvin. Also note that 1atm is approximately [itex]1.01\times 10^5[/itex] pascals.

Reagards,
-Hoot
Thanks!!

so for 1 liter, I get 1.0 x 10^-3 m^3

"1.0 times ten to the minus three meters cubed"

Using the equation PV=nRT

and solve for P

P={(2.00 moles)(8.314 J/mol)(-50 dgrees Celsius + 273K)/1.0 x 10^-3 m^3}

P={(2.00)(8.314)(223K)/(.001 meters cubed)

P=3708044 (what is the unit for this answer...is it millimeters or ?)

So to obtain the correct pressure (in atm), I divided my answer of 3708044 by (1 x 10^5 pascals) to get approximately 37.08044 atm or 37.1 atm??

Am I close?? LOL :smile:
 
  • #8
Hootenanny
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thunder said:
So to obtain the correct pressure (in atm), I divided my answer of 3708044 by (1 x 10^5 pascals) to get approximately 37.08044 atm or 37.1 atm??

Am I close?? LOL :smile:

Looks right to me :smile:

Regards,
-Hoot
 
  • #9
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Hootenanny said:
Not quite. As the gradient is given in atm L / mol, you need to use that form of the universal gas constant (i.e 0.0821). Also the T in the equation represents absolute temperature, so there's no need to add 273 to your answer.

Regards,
-Hoot
LOL, oops, ok thanks. Let's see then what I need to do is as follows:

nRT=45.6 atm

replace n with 2.00 moles
repalce R with .0821 atm L/mole

and solve for T

(2.00)(.0821)T=45.6

DIVIDE (2.00)(.0821) OR (.1642) from both sides

and solve for T

T=45.6/.1642

T=277.710109622 Kelvin OR 277.7 Kelvin

Correct?? :smile:
 
  • #10
Hootenanny
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thunder said:
(2.00)(.0821)T=45.6

Almost, you've used 2 moles when it should be 3, I think you've confused it with your second question :smile:

Regards,
-Hoot
 
  • #11
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Hootenanny said:
Looks right to me :smile:

Regards,
-Hoot
Yippeee!!!!! Thanks!!! Ok so that takes care of problem #2.

I redid problem # 1, where I am looking for the correct Kelvin temperature of the gas (see my last post above). Am I close? How'd I do? Really appreciate you helping me work thrugh this to find the problem.
 
  • #12
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Hootenanny said:
Almost, you've used 2 moles when it should be 3, I think you've confused it with your second question :smile:

Regards,
-Hoot

oops!! LOL

Yep, I'll correct that!! Thanks!! :smile:
 
  • #13
Hootenanny
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I'm talking about you re-do for question one. You've used 2 moles when it should be 3!!
 
  • #14
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Hootenanny said:
I'm talking about you re-do for question one. You've used 2 moles when it should be 3!!
ok , gotcha...

I reworked it as follows:

nRT=45.6 atm L

n=3.00 moles :smile:
R=.0821
and solve for T

(3.00 moles)(.0821)(T)=45.6

DIVIDE both sides by (3.00)(.0821) OR (.2463)

and solve for T

T=45.6/.2463

T=185.140073082 Kelvin or approximately 185 Kelvin


So, as Regis would say, my final answer for problem # 1 is approximately 185 Kelvin.

And for problem #2, I got approximately 36.7 atm.

Whewww!!! Correct?? o:)
 
  • #15
Hootenanny
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They look correct to me, they're the same as what I got anyway :biggrin:

Well done.

Regards,
-Hoot
 
  • #16
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Hootenanny said:
They look correct to me, they're the same as what I got anyway :biggrin:

Well done.

Regards,
-Hoot
Awesome!! Thanks so much for you assistance! Really appreciate it! BTW, I love that Einstein quote!! :biggrin:
 
  • #17
Hootenanny
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thunder said:
Awesome!! Thanks so much for you assistance! Really appreciate it!

My pleasure
 

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