Ideal Gas Law

[Richay]

The volume of a quantity of the ideal gas was kept constant in a experiment. The final temp. was 600 kelvins (K) and the final pressure was 300 netwons per square m (Meter). What was the original pressure if the original temperature was 1000 K?


This is so mind twisting.

I got it wrong three times

[Ouabache]

What was your approach on this question? Also please be sure to re-read this thread (http://www.physicsforums.com/showthread.php?t=94379) part #1.

If you show us what you have done, we can help steer you in a successful direction.

What kind of relationship (equations) do you know that relate the variables in your question?

[Richay]

K, I read it.

I was doing the problem wrong. But I think i understand it now.
My equation is

1. 1000 x 300 = 300,000
2. 300,000 / 600 = 500

Answer=500

This all works out right?

[Ouabache]

I don't understand.. What are you trying to accomplish with your calculations? Do you know the equation that shows the relationship between your variables (final temperature, pressure, & volume; original temperature, pressure & volume)?

[Richay]

I'm trying to accomplish what the original pressure was if the original temperature was 1000 K.
1000 [Original Temperature] x 300 [Final Pressure] = 300,000

300,000 [New Solution] / 600 [Final Temperature] = 500 [Original Pressure]

Answer = 500

I asked my advisor if i was correct, and he said yes.

But, I want to know if you think my steps were the correct way of getting the answer?

[Ouabache]

Yes your solution is okay..

It helps to understand the relationship of your variables to solve any problem. That way you are not limited by step-by-step instructions shown in your text or given by the instructor.

In your case one form of the general expression for The Ideal Gas Law, where i is initial and f is final, P-pressure, V-volume and T-temperature.

\frac {P_i V_i}{T_i} = \frac {P_f V_f}{T_f}

ref (http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/idegas.html) (see Ideal Gas Law with Constraints)

Since your volume is constant V_i = V_f

You are asked what is the original pressure. Solving for P_i
(using basic algebra)

P_i = \frac {P_f T_i}{T_f}
Once you realize that, the solution becomes trivial (just plugging in the numbers and keeping your units consistent). Speaking of units, I hope remembered to give your solution in \frac {N}{m^2}