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Work done by gas and balloon?

  • #1
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Homework Statement


The question is:
One kilogram of a gas with a molecular weight of 35 is contained in a balloon. The initial conditions are 27 C and 0.025 m^3. The gas is slowly heated isothermally until the final volume is 0.05m^3. The atmospheric conditions are 100 kPa and 27 C. The p-v-T relationship for the gas is given by pv = RT(1+bv) where R = 8.135 kpa-m^3/(kg mol-K) and b = 0.012 kg/m^3. Calculate the work done during the process for (a) the gas and (b) the gas and balloon.


Homework Equations


pv = RT(1+bv) (for the gas p-v-T relation). Other equations may be molecular weight = g/mol? Conversions to K, the work equation (W = ∫pdv).

The Attempt at a Solution


With the info given I can calculate the an initial pressure and final pressure of the balloon (I think?), so for work done, its just W = p2v2 - p1v1? and for the gas, do I just integrate P, where P = RT(1+bv)/v ? When trying this integration for work done, the equation is dependent on the volume, so I'm a bit confused on if this is the work done by the balloon or the gas. The result I got from this method using integration was W = 1692.358 kJ (so probably gas?). Any guidance/suggestions would be appreciated!
 
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Answers and Replies

  • #2
Charles Link
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If you compute the internal pressure of the balloon, (as a function of volume), you should find, with the equation of state they give you, that the pressure is always greater inside the balloon than it is outside the balloon, (at least for volumes from .025 to .050 m^3). The work done by the inside gas plus balloon system on the outside atmospheric pressure, (part (b)), can be readily computed for the expansion, and it will be less than the ## \int P_{internal} \, dV ##, which is the work done by the gas inside the balloon, (part (a)). ## \\ ## I have one problem with the above though: How do you heat something isothermally ? That makes no sense. If they are saying they softened up the balloon material, by applying a little heat that kept the gas inside the balloon at about the same temperature = kept the gas basically at temperature ## T=27 \, C ## , so that they have a larger final volume at the same temperature and thereby a lower final pressure, it might make sense. Otherwise, I think the problem is simply in error. ## \\ ## Also note: Their ## v=V/n ##. ## \\ ## Also, a rather inaccurate ## R ## with incorrect units: ## R=.08206 ## liter-atm/(mole K)=##8.314 \, ## spa m^3/(mole K). There is no "kg" in the denominator in the units. And the "spa-m^3"=1 Joule. (spa=standard pascal=1 N/m^2). I think the problem is rather lacking in overall quality. ## \\ ## The more I look at this, the more unrealistic this problem is. The balloon is approximately 1 ft^3 in volume initially. It could not contain 1.0 kg of gas= 2.2 pounds of gas. That is simply impossible.
 
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  • #3
kimbyd
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I have one problem with the above though: How do you heat something isothermally ?
That threw me a little at first too, but then I realized that heating, in a thermodynamic sense, just means inputting energy. It's totally possible to input energy and not increase the temperature. Heck, self-gravitating gas clouds (as you work with in astrophyics) tend to drop in temperature if heated, or increase in temperature if cooled. This is why the nuclear fire at the center of a star actually causes the star to stop heating up: if it weren't for the large input of energy from the thermonuclear reaction, the star would just keep collapsing under its own gravity, heating up all the way until it became supported by degeneracy pressure.

With the balloon, the expansion has to be enough that the heat input into the system all goes to causing the balloon to expand, rather than increasing its temperature.
 
  • #4
Charles Link
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That threw me a little at first too, but then I realized that heating, in a thermodynamic sense, just means inputting energy. It's totally possible to input energy and not increase the temperature. Heck, self-gravitating gas clouds (as you work with in astrophyics) tend to drop in temperature if heated, or increase in temperature if cooled. This is why the nuclear fire at the center of a star actually causes the star to stop heating up: if it weren't for the large input of energy from the thermonuclear reaction, the star would just keep collapsing under its own gravity, heating up all the way until it became supported by degeneracy pressure.

With the balloon, the expansion has to be enough that the heat input into the system all goes to causing the balloon to expand, rather than increasing its temperature.
Please study the complete problem more carefully, as well as my responses to it. I think you will conclude the problem is completely incorrect.
 
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  • #5
kimbyd
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Please study the problem more carefully, as well as my response to it. I think you will conclude the problem is completely incorrect.
I definitely believe that the problem is unphysical (it would require a gas which has a negative heat capacity, which to my knowledge doesn't happen unless self-gravity is significant). The system also requires the pressure of the gas inside the balloon to drop as the gas inside it expands. That seems unlikely.

There's also an error in the units for R and b, but it looks like it should be pretty easy to calculate a result if we assume the units were just written incorrectly.

The work the gas does is just:

$$W = \int_{v_0}^{v_1} p(v) dv$$

To get ##p(v)## you can use the provided gas law (with T = const). The integral shouldn't be too difficult.

Then, the work the gas/balloon system does on its surroundings is even simpler, as the outside pressure is a constant 125kPa, making the integral trivial.

The difference in work between the two would represent the change in energy of the balloon itself.

This problem is definitely wonky, and the result definitely won't make much sense. I strongly suspect that this is a problem that was either poorly thought-out from the start, or was modified poorly from an initial question that made sense. But I'm pretty sure it has a well-defined solution.
 
  • #6
Charles Link
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Please look at the last line in my response, where I finally looked at the weight of the gas and the volume it occupies.
 
  • #7
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If you compute the internal pressure of the balloon, (as a function of volume), you should find, with the equation of state they give you, that the pressure is always greater inside the balloon than it is outside the balloon, (at least for volumes from .025 to .050 m^3). The work done by the inside gas plus balloon system on the outside atmospheric pressure, (part (b)), can be readily computed for the expansion, and it will be less than the ## \int P_{internal} \, dV ##, which is the work done by the gas inside the balloon, (part (a)). ## \\ ## I have one problem with the above though: How do you heat something isothermally ? That makes no sense. If they are saying they softened up the balloon material, by applying a little heat that kept the gas inside the balloon at about the same temperature = kept the gas basically at temperature ## T=27 \, C ## , so that they have a larger final volume at the same temperature and thereby a lower final pressure, it might make sense. Otherwise, I think the problem is simply in error. ## \\ ## Also note: Their ## v=V/n ##. ## \\ ## Also, a rather inaccurate ## R ## with incorrect units: ## R=.08206 ## liter-atm/(mole K)=##8.314 \, ## spa m^3/(mole K). There is no "kg" in the denominator in the units. And the "spa-m^3"=1 Joule. (spa=standard pascal=1 N/m^2). I think the problem is rather lacking in overall quality. ## \\ ## The more I look at this, the more unrealistic this problem is. The balloon is approximately 1 ft^3 in volume initially. It could not contain 1.0 kg of gas= 2.2 pounds of gas. That is simply impossible.
My apologies! Should have been looking for the replies! Its meant to be kpa I think (which might help with some confusion), not spa, the problem sheet had a few typos and I typed it as it was written and missed it. Also, I am also a little confused by 'heating' (unless they just mean energy, like someone was saying) if the temperature is constant. Maybe we are supposed to assume that amount of gas is being 'put' into the balloon? This problem's wording is definitely wonky, (and I think numbers were just randomly chosen, so not a good design as you are saying). I'll respond tomorrow after I ask about it in class to clear some things up.
 
  • #8
Charles Link
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What would make more sense would be a mass of about 80 to 100 grams in the balloon. Air (M.W. ## \approx ## 30) , has a mass of 1.28 kg/m^3 at STP. To have 1 kg=1000 g in a balloon volume of .025 m^3 is simply unrealistic. ## \\ ## Also, I gave a possible explanation if the problem were worked isothermally. The balloon would also need to get more flexible as it expanded, which is not completely unrealistic, but a little out of the ordinary. At lower internal pressure, it would need to have a larger volume. Unless the balloon material underwent a change, which wasn't specified in the problem, it simply wouldn't expand like that.
 
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  • #9
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Also, I gave a possible explanation if the problem were worked isothermally.
Ah okay I see, thats what my calculations showed, that p2 < p1 (as it should for what was given). But like you said, the problem needs more clarification. The assumptions that to be made for the problem to make physical sense due to the lack of information are a little confusing. I'll for sure bring this up in class. Thanks for the replies!
 
  • #10
Charles Link
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I should add a little about how an ordinary balloon is made to expand, particularly if you assume the temperature stays the same: The pressure is normally just slightly above 1.0 atm inside the balloon, maybe 1.10 or 1.15 atm. If you double the volume at constant temperature, it is usually accomplished by doubling the number of air molecules ## n ## in the balloon, (using ## PV=nRT ##), and perhaps a little bit more, because an increase in volume may require a slight increase in pressure, e.g. from 1.10 atm to 1.12 atm. ## \\ ## [Note: I think sometimes the pressure in a balloon actually decreases slightly with an increase in volume=Edit: it is unrealistic, as the balloon in the above problem behaves, to have the absolute pressure drop by a factor of 2 as the balloon doubles in size=the gauge pressure (absolute pressure minus atmospheric pressure) might behave like this........ (e.g. the (absolute) pressure might go from 1.15 atm to 1.075 atm as the volume doubles, so that the gauge pressure goes from .15 to .075 atm=and when you are blowing it up you might recognize this decrease in pressure if you were blowing into it), ## \\ ## .... but not the absolute pressure]. ## \\ ## And one additional item: The units on ## b ## should be moles/m^3. ## \\ ## Additional item: To inflate something like a basketball to 5 or 6 psi( lbs./in^2) of pressure above atmospheric pressure (1.0 atm=14.7 lbs./in^2), so that ## P_{basketball} \approx 1.40 ## atm requires a tire pump or a hand pump. Something like an ordinary balloon or a beach ball is at a lower pressure=my estimates are 1.10 to 1.15 atm, and I think a google would confirm this. ## \\ ## Edit: And a google shows the average person can generate about 2 psi (lbs/in^2) ## \approx ## .15 atm with their lungs, so my estimates are reasonable.
 
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  • #11
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The v in the equation of state is obviously molar volume in units of m^3/kg-mol (as pointed out by Charles Link). Are you sure about the units of b, or should it more properly be 0.012 kg-mol/m^3?

The initial value of v is 0.025/(1/35) = 0.875 m^3/kg-mol. So, assuming b = 0.012 kg-mol/m^3, bv = 0.0105. And the initial pressure in the balloon is (8.314)(300)(1.0105)/(0.875)=2880 kPa = 28.8 bar. Sounds high, but the gas is pretty dense. This is practically identical to what the ideal gas law predicts.
 
  • #12
Charles Link
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The v in the equation of state is obviously molar volume in units of m^3/kg-mol (as pointed out by Charles Link). Are you sure about the units of b, or should it more properly be 0.012 kg-mol/m^3?
Scratch that=I made a mistake. (I forgot that the initial mass is 1.0 kg=1000 g, making the number of moles ## n=1000/35 ##, so that ## v=\frac{V}{n}=.875 \cdot 10^{-3} ## m^3/(gram-mole)=.875 m^3/(kg-mole), as @Chestermiller has correctly computed above).## \\ ## It would make sense, (as @Chestermiller has mentioned), that ## b=.012 ## kg-mole/(m^3) , rather than having ## b= .012 ## gram-mole/m^3=.000012 kg-mole/(m^3). ## \\ ## And I see the OP is now edited so that ## R ## reads correctly as ## R=8.135 ## kpa-m^3/(kg-mole K) with the "spa" changed to a "kpa". (More accurately it would be ## R=8.314 ## with the same units). ## \\ ## And I still think that P= 28 atmospheres (with 1.0 kg of gas in a volume of 25 liters) is quite a considerable pressure for a balloon to have. And you normally wouldn't get the balloon to expand from .025 m^3 to .050 m^3 by heating it "isothermally". The volumes chosen seem to be very typical numbers for an ordinary balloon, but much of the rest of the problem seems to be somewhat out of the ordinary.
 
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