# How Do You Calculate Temperature and Pressure in Boyle's Law Problems?

• thunder
In summary: I get it this time? :smile:And just to be sure, I only used 1-2 significant figures throughout this problem, right? I think my problem #1 I used too many sig figs...but I'll live with that! LOL
thunder
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!

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!

Well, you can use your equation;

$$P = \frac{nRT}{V}$$

Now consider the general equation of a straight line;

$$y =mx +c$$

$$P = nRT \cdot \frac{1}{V}$$

Regards,
-Hoot

 I've just seen your other post now thunder, welcome to PF![/Edit]

Last edited:
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! :)

For your second question, you need only apply the equation

$$pV = nRT$$

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

Reagards,
-Hoot

Hootenanny said:
Well, you can use your equation;

$$P = \frac{nRT}{V}$$

Now consider the general equation of a straight line;

$$y =mx +c$$

$$P = nRT \cdot \frac{1}{V}$$

Regards,
-Hoot

 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.

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

Hootenanny said:
For your second question, you need only apply the equation

$$pV = nRT$$

Ensure that the volume is in $m^3$ and the temperature is in kelvin. Also note that 1atm is approximately $1.01\times 10^5$ 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

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

Looks right to me

Regards,
-Hoot

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??

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

Regards,
-Hoot

Hootenanny said:
Looks right to me

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.

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

Regards,
-Hoot

oops! LOL

Yep, I'll correct that! Thanks!

I'm talking about you re-do for question one. You've used 2 moles when it should be 3!

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
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??

They look correct to me, they're the same as what I got anyway

Well done.

Regards,
-Hoot

Hootenanny said:
They look correct to me, they're the same as what I got anyway

Well done.

Regards,
-Hoot
Awesome! Thanks so much for you assistance! Really appreciate it! BTW, I love that Einstein quote!

thunder said:
Awesome! Thanks so much for you assistance! Really appreciate it!

My pleasure

## What is Boyle's Law?

Boyle's Law is a gas law that describes the relationship between the pressure and volume of a gas at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume, meaning that as the volume increases, the pressure decreases and vice versa.

## Who discovered Boyle's Law?

Boyle's Law was discovered by the Irish natural philosopher and chemist, Robert Boyle, in the 17th century. He conducted experiments on the properties of gases and published his findings in his famous work, "The Sceptical Chymist" in 1661.

## What is the formula for Boyle's Law?

The formula for Boyle's Law is P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume of a gas, and P2 and V2 represent the final pressure and volume of the gas at a constant temperature.

## What are some real-life applications of Boyle's Law?

Boyle's Law has many practical applications, such as in scuba diving, where the pressure of compressed air in a tank decreases as the diver ascends to the surface, following the principles of Boyle's Law. It is also used in the functioning of gas-powered engines and in the production of industrial gases.

## What are the limitations of Boyle's Law?

Boyle's Law is only applicable to ideal gases at a constant temperature. It does not take into account factors such as intermolecular forces and gas particle size, which can affect the behavior of real gases. Additionally, it only applies to a closed system where the amount of gas remains constant.

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