How many moles of air are in the tank?

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Homework Help Overview

The discussion revolves around determining the number of moles of air in a compressed-air tank, given its dimensions, pressure, and temperature. The subject area includes gas laws and volume calculations in the context of ideal gas behavior.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the ideal gas equation and its application to the problem. There are attempts to calculate the volume of the tank and the number of moles of air, with some questioning the correctness of their volume calculations and unit conversions.

Discussion Status

Participants are actively engaging with the problem, providing hints and corrections regarding the volume calculations and unit conversions. Some have noted discrepancies in their results and are seeking clarification on the ideal gas law application.

Contextual Notes

There is a focus on ensuring correct unit conversions, particularly between cm³ and m³, and the appropriate use of the gas constant. Participants are also discussing the implications of using different pressure units at standard temperature and pressure (STP).

mawalker
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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?
 
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Do you know the ideal gas equation? That should be all you need for both parts.
 
yes, pV = nRT... would the volume be 4/3 pi(7)^3*(40) = 57470.2 ?
 
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.
 
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...
 
do i need to do an additional conversion to get volume into L?
 
I'd just convert atmospheres into pascals 1atm = 1.01x10^5 Pa, and use the standard form of the gas constant.
 
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
 
Your volume should be in m^3.
 
  • #10
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}
 
  • #11
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
 
  • #12
1 cm^3 = (1/100)^3 m^3 = 1E-06 m^3
 
  • #13
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.
 
  • #14
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.
 
  • #15
thanks, it was a silly error...
 

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