1. Mar 4, 2009

### alex-book

1. The problem statement, all variables and given/known data

Hi I am an Engineering student and i have to do an analysis on a closed system vessel that is going to be heated up.And before i start the experiment i want to an analysis or a safety factor for this experiment.

I am going to have a closed system vessel, lets say the volume is going to be 3 m^3. and surface area of 3m^2.
Initially i am going to put an argon gas inside the closed vessel when the vessel is still at room temperature (25C) and then heat it up until it reach 500Celsius.(T=773K)

2. Relevant equations

Pv=nRT, and density=m/V

3. The attempt at a solution

and gas volume is always the same with the medium right? so the argon gas volume on a 3m^3 vessel will be 3m^3 as well right?

but isnt it that the volume of the gas depends on the temperature and the pressure?

or can i just calculate the mass by using the density of the argon gas at 25C and find the mass or the mass is going to be different at 500C?

Sorry about a lot of confusion in my posting, i am really confused and do need help. Thank you

2. Mar 4, 2009

### Redbelly98

Staff Emeritus
Actually a 3 m^3 volume must have a surface area of at least 10 m^2, but that isn't really important for this problem.

Yes.
Those 3 are all related to each other, through PV=nRT.

Since V is a constant (in this situation), what can you say about the ratio P/T?

p.s welcome to PF

3. Mar 4, 2009

### alex-book

Thnx redbelly98!

ohhh so the volume stays the same? well :) i kno its a stupid question, its just that the volume tend to increase with the temperature right?
And it is true right for Argon gas i can just use ideal gas equation, or should i go with the non-ideal gas formula?

Thanks again!

4. Mar 5, 2009

### Redbelly98

Staff Emeritus
At room temperature and 1 atm pressure, argon is close to an ideal gas.

To calculate the final pressure, you can try both ideal and real gas solutions and compare them.

Since n/V is fixed, and can be calculated at room temp & 1 atm, the real gas calculation is not as complicated as you might think.

5. Mar 5, 2009

### alex-book

Yea i have tried both real gas and ideal gas equation and both of them almost the same, took me so long to find the van der waals constant+equation. Thanks Redbelly!
and for both of calculation i used the density of argon at STP(1atm and 25C) which is 1.449kg/m3 is that okay?

so it will be like this

PV=nRT --> n = m/M and Rspecific = R/M so PV=mRspecific T --> P = density*Rspecific*T

so P = density * Rspecific * Temp
= 1.449 kg/m3 *0.208 KJ/kgK *773K
= 232.2 KPa = 2.29 atm (I have tried to calculate it using Vdwaals equation and it got me almost the same result. )

is that correct?
but it is just weird to use the density of STP value to calculate in 500C temperature. Because i know in h2O there is different value of density that you have to use in different type of temperature and pressure, this confuses me, like do i need to find the exact value of density of argon at 500C or i can just use the SPT one?
but in the other hand density is mass/volume and in this case mass and volume stays the same since its a closed vessel, so may be i could use the same density value, is that the reason?
I am sorry if i confuse you with my 100000questions, is just that i want to learn+do this right :)

Last edited: Mar 5, 2009
6. Mar 5, 2009

### Redbelly98

Staff Emeritus
You're making it more complicated than necessary.

PV = nRT

Volume and # of moles are constant. So what is the relation between P and T?

7. Mar 5, 2009

### alex-book

directly proportional?
k thanks redbelly! i think i got it! thnx

8. Mar 5, 2009

### Redbelly98

Staff Emeritus
Yes, exactly.
You're welcome