How many moles of air are in the tank? (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

mawalker

I am having trouble setting up this problem. Especially part A of the problem. I think I have a good idea of what to do for part B, but could anyone give me any hints on how to find moles from the given information?

A 14.0 cm-diameter compressed-air tank is 40.0 cm tall. The pressure at 30.0 C is 150 atm.

A.

How many moles of air are in the tank?

B.

What volume would this air occupy at STP?

Tomsk

Do you know the ideal gas equation? That should be all you need for both parts.

mawalker

yes, pV = nRT.... would the volume be 4/3 pi(7)^3*(40) = 57470.2 ?

Tomsk

No, if it's a cylinder the volume is $2\pi r^2 h$.

$\frac{4}{3}\pi r^3$ is the volume of a sphere.

mawalker

thanks, i always get the volume equations mixed up. using pV=nRT i got 150 atmospheres(pressure)*12315.04L(Volume)= n * .08206 (gas constant) * 303 (temperature K)

which gave me 1847256 = 24.86n

solving for n i got n = 74293.86.... this seems like a lot of moles....

mawalker

do i need to do an additional conversion to get volume into L?

Max Eilerson

I'd just convert atmospheres into pascals 1atm = 1.01x10^5 Pa, and use the standard form of the gas constant.

mawalker

hmmm... that made it even worse... i took (15150000 pa)(12315.04 cm3) = n (8.3145) (303) solving for n gave me 74057610.2 ... this still seems like a lot of moles

Max Eilerson

Your volume should be in m^3.

Max Eilerson

Just noticed your volume is out by a factor of two anyway, since the volume of a cylinder is $$\pi r^2 h = \frac{\pi d^2 h}{4}$$

mawalker

ahhh, i'm still getting a ton of moles..... converting from cm^3 to m^3 i got 61.58 m^3....

so 15150000 pa (61.58 m^3) = 93286420

(8.3145) constant * 303 degrees K = 2519.29n

divide both sides by 2519.29

and i find that n = 370288

Max Eilerson

1 cm^3 = (1/100)^3 m^3 = 1E-06 m^3

mawalker

ahhh, thank you....that was the problem... ok now i'm having a little bit of trouble with part B. it asks how much volume it would occupy at stp

at stp i know:
T= 0 C or 273 K
P=1 atm or 1.01 *10^5 pa

so using pV=nRt

(1.01*10^5)V = (37.1 mol)(273 K)(8.3145 constant)

solving for V i get 3.08 m3 and this answer is not correct. i don't see what i am doing wrong here.

Max Eilerson

Isn't standard pressure 1bar, i.e. 1x10^5 Pa not that it makes a difference.
Rearrange the equation before you put the numbers is in it makes it much easier. V = nRT / P, I assume you're making some silly mistake because the numbers look correct.

mawalker

thanks, it was a silly error...

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving