# Homework Help: Ideal Gas Law

1. May 2, 2006

### 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

2. May 2, 2006

### Ouabache

What was your approach on this question? Also please be sure to re-read this thread 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?

Last edited: May 2, 2006
3. May 3, 2006

### 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

This all works out right?

4. May 3, 2006

### 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)?

5. May 4, 2006

### 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]

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?

6. May 4, 2006

### 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 (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 $$\frac {N}{m^2}$$

Last edited: May 4, 2006