Ideal Gas Law and Balloon Volume

In summary: The pressure inside the balloon is 1 m3. I just assumed that.Suppose the volume of a balloon decreases so that the temperature of the balloon decreases from 280K to 240K and its pressure drops from 1.6x10^4 Pa to 1.7x10^4 Pa. What is the new volume of the gas.The new volume is 1.7x10^4 m3.
  • #1
Virmeretrix
6
0
Hey guy,
I didnt think i would ever have to deal with gas laws after i got out of AP chem, but it has showed up again in physics. Tell me what to do.

Suppose the volume of a balloon decreases so that the temperature of the balloon decreases from 280K to 240K and its pressure drops from 1.6x10^4 Pa to 1.7x10^4 Pa. What is the new volume of the gas.

My question is what was the initial? How do i work this prob.? Thanks for any help.
 
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  • #2
Let V1 be the unknown initial volume. Let V2 be the unknown final volume. At this point, I am not sure exactly what equation you are using. I assume the following:

PV = nRT, where n = number of moles, R = universal gas constant, P = pressure, and V = volume.

It does not seem to me that you need to know the initial volume. Just solve for V = nRT/P and put in the final values for pressure and temperature. Or am I missing something or you left something out?
 
  • #3
I need to know the New Volume of the gas. We have to use PV=NKbT where Kb is 1.38x10^-23. I was trying to use P1V1/T1=P2V2/T2. I just don't know where to start.
 
  • #4
Originally posted by outandbeyond2004
It does not seem to me that you need to know the initial volume. Just solve for V = nRT/P and put in the final values for pressure and temperature. Or am I missing something or you left something out?

The problem is that he doesn't know what "n" is. He has 2 variables and only 1 equation.

To solve the problem, I'm guessing you would need a formula for n or V that does not have the other in it. Then you could solve 2 variables with 2 equations.
 
  • #5
The ideal gas law says the volume of an ideal gas is proportional to its temperature and inversely proportional to its pressure. So when you know the initial volume V1 and the fraction by which the temperature increases T2/T1 you can calculate the final voulume: V2=V1(T2/T1). Considering a simultanuous change of pressure from p1 to p2 the final volume can be calculated to be V2=V1(T2/T1)(p1/p2).
 
  • #6
Originally posted by da_willem
the final volume can be calculated to be V2=V1(T2/T1)(p1/p2).

He said in his first post that he doesn't know what V1 is.
 
  • #7
Originally posted by ShawnD
He said in his first post that he doesn't know what V1 is.
da_willem's answer is the best you can do with the given information.
 
  • #8
P1V1 = nKbT1
P2V2 = nKbT2

These are 2 equations with 3 unknowns, no? Need another independent equation. I suspect it involves n for air.

Or, maybe all you need is the RATIO of final volume to initial volume:

V2/V1 = (P1/P2)(T2/T1)
 
  • #9
Hey guys... uh, i really didnt think the prob. would be soo difficult. Let me tell you my reasoning behind the way i worked it, and you tell me if you think it is ok.

#1. The pressure inside the balloon is 1 m3. I just assumed that.
#2. I plugged that into the P1V1/T1=P2V2/T2 equation, which gives me that V2=.807 m3 i believe, i don't have my work in front of me.

Does that sound good?
 
  • #10
I ment that the volume is 1 m3 not the pressure
 
  • #11
PV=nRT, the n remains the same.

so V=RT/P.

In each case, before and after, we can find V since we know both T and P. So

Vinitial=RTinitial/Pinitial

Vnew=RTnew/Pnew
 
  • #12
Originally posted by GeneralChemTutor
PV=nRT, the n remains the same.

so V=RT/P.
...
No. While the n remains the same, since you don't know what it is you can't use it to find V. (And you can't just drop it from the equation.) The best you can do is find the final volume in terms of the initial volume.

Unless, of course, you are willing to just assume a value for n or for the initial volume. (Like Virmeretrix did when he arbitrarily set the initial volume equal to 1 cubic meter--why?)
 
  • #13
I don't know what Viry's textbook says, but "one gram-mole of an ideal gas occupies a volume of 22,400 cm3 or 22.4 liters at "standard conditions" or at 'normal temperature and pressure' (NTP), that is, at a temperature of 0degreesC = 273degreesK ((sic)) and a pressure of 1atm = . . ."

The first example in my textbook for the ideal gas law involved oxygen, and treated it as tho it were an ideal gas.

I believe Viry now has everything he needs to solve the problem.
 
  • #14
Ok got the answer.

GeneralChemTutor is right.
The n, which in this case is the molecules in the balloon, remains constant sense the balloon is a closed system. Once again, we were not to use R, the universial gas constant, but rather Kb, know as the bezionne or something constant.
Sense the question did not state weather or not the system was at STP, you have to assume its not.
So, [P1V1=N1Kb1T2]=[P2V2=N2Kb2T2]. Sense the balloon is a closed system the n(number of moles/atoms/molecules) no matter what the numerical value, devides out. So in each case, V1=KbT1/P1 which would give you the initial volume to be 2.4x10^-25 m3. and V2=KbT2/P2 gives you 1.9x10^-25 m3.
Pretty small balloon huh?
 
  • #15
Uh, oh, does the -trix ending mean Viry is a female? If so, I apologize.

No, you still need to know n or N. The answers you got is for N = 1. One atom, I think? That would be a small balloon, all right.
 
  • #16
Actually, I'm a male. The name is a combination of two latin words. Vir- meaning man, and meretrix, meaning W*ore. So, put the two together. And you say I am still wrong. Then what do you suggest?
 
  • #17
It's clear by now that either you copied the question wrong, or forgot to include vital information. Whether you use R or Kb, you still need to know n or N, which you don't. As has already been said, the most you can find is the ratio between the old volume and the new volume, or the new volume as a function n or N.
 
  • #18
Virmeretrix said:
So, [P1V1=N1Kb1T2]=[P2V2=N2Kb2T2]. Sense the balloon is a closed system the n(number of moles/atoms/molecules) no matter what the numerical value, devides out.
That is incorrent. Ignoring for a second that P1V1 does not equal P2V2 or N2Kb2T2, you cannot say that because:
2x = 2 = y = 2
You can divide only two parts of the equation by 2, to get:
2x = 1 = y = 1
If you want to divide by 2, you must divide all parts:
x = 1 = y/2 = 1
 

Related to Ideal Gas Law and Balloon Volume

1. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is represented by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

2. How does the Ideal Gas Law apply to a balloon?

In a balloon, the Ideal Gas Law is used to predict the behavior of the gas inside the balloon. As the temperature of the gas increases, the volume of the balloon will also increase, causing the balloon to expand. Similarly, if the temperature decreases, the volume of the balloon will decrease, causing it to shrink.

3. What is the significance of the gas constant in the Ideal Gas Law?

The gas constant, represented by the symbol R, is a proportionality constant that relates the pressure, volume, and temperature of an ideal gas. It is a universal constant that is the same for all gases, regardless of their properties, and is used to convert units in the Ideal Gas Law equation.

4. How do changes in pressure and temperature affect a balloon filled with an ideal gas?

As per the Ideal Gas Law, an increase in pressure will result in a decrease in the volume of the balloon. Similarly, an increase in temperature will cause the balloon to expand. This is because when pressure is applied to the gas molecules, they will move closer together, thus occupying a smaller volume. On the other hand, when the temperature increases, the gas molecules will move faster and take up a larger volume.

5. Can the Ideal Gas Law be applied to non-ideal gases?

The Ideal Gas Law is an approximation and is most accurate for ideal gases, which follow the assumptions of the kinetic theory of gases. However, it can also be applied to non-ideal gases under certain conditions, such as high temperatures and low pressures. In these cases, corrections are made to the ideal gas law equation to account for the deviations from ideal behavior.

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