# 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