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Topic: Pressure & Temperature at a constant volume  (Read 3307 times)

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Offline whippedswalbr

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Pressure & Temperature at a constant volume
« on: March 20, 2010, 04:05:43 AM »
I haven't got a clue how to do this when I don't have V,moles. (PV=nRT)
A gas cannister, in which the pressure is initially 1.20 atm at 10°C, is heated until the internal pressure is 2.62 atm.
What will be the new temperature, in °C, inside the canister?

Offline Schrödinger

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Re: Pressure & Temperature at a constant volume
« Reply #1 on: March 20, 2010, 05:21:39 AM »
Read the question again ...carefully.
What are the changes that are taking place in the system?
Now, can you reason out why you weren't given 'V' and 'n'?
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Offline billnotgatez

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Re: Pressure & Temperature at a constant volume
« Reply #2 on: March 27, 2010, 10:53:53 AM »
Since this topic was posted a while ago, I guess I can use it without doing homework for someone.
Anyone can correct me if you see issues.

When analyzing the problem in general it looks like you can look at it as an ideal gas in situation 1 and situation 2.

Mathematically it can be represented by
Situation 1 ---- P1V1= n1RT1
Situation 2 ---- P2V2= n2RT2

It follows mathematically

Situation 1 ---- R = (P1V1) / (n1T1)
Situation 2 ---- R = (P2V2) / (n2T2)

Since R is the same in both situations

(P1V1) / (n1T1)= (P2V2) / (n2T2)

Now for the problem at hand

We know that since we have a closed container the volume has to stay the same for both situations and if there are no leeks the number of molecules will be the same for both situations.
So using these situations we have V and n canceling out of the equation.

Therefore we can use the following to solve for T2
(P1) / (T1)= (P2) / (T2)
Or
 (T2) = (T1) (P2) / (P1)

When doing ideal gas law calculations we need to make sure all the values have the same units and are absolute.
So the temperatures has to be converted to K (Kelvin) instead of C (centigrade).
In this case I will use 273 K to convert from 10 C to 283 K [sometimes 273.15 is used depending on interest in accuracy].

So plugging in the values

T2 = (283K x 2.62ATM) / (1.2ATM)
T2 = 617.88333333333333333333333333333 K
I guess we should change the answer based on significance.
Here is where I am not sure what to do
Is it 617.88 K or 617.9 K or 618 K
Which is 344.88 C or 344.9 C or 345 C

Hopefully I did all the math correctly



Offline Borek

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Re: Pressure & Temperature at a constant volume
« Reply #3 on: March 27, 2010, 11:52:33 AM »
(P1) / (T1)= (P2) / (T2)

http://en.wikipedia.org/wiki/Gay-Lussac's_Law

Quote
Is it 617.88 K or 617.9 K or 618 K

I would go for 618.
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Offline billnotgatez

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Re: Pressure & Temperature at a constant volume
« Reply #4 on: March 27, 2010, 01:26:18 PM »
Ah - yes, I should have posted

 (P1)/(T1) = (P2)/(T2)
Or
 (P1T2) = (P2T1)
Or
 (T2) = (T1 P2)/(P1)


Yes 618 K seems best

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