# Helium volume exam questions

1. Nov 25, 2003

### Security

The two questions that were on my exam, to be honest, i didnt answer because I didnt understand it enough to do so. But Im willing to learn and be shown the way to a legit answer, so please dont immediately give me answers.

The questions are like this:

The density of air near the earth's surface is 1.29kg/m3. If a helium ballon with a mass of 1 kg floats in air without rising of falling, what is the minimum volume of helium in the ballon? Show your calculations.(Presume that the mass of the material making up the balloon is negligible.)

-- and --

A balloon filled with one liter of air is tied to a brick and dropped from a height of 4600 m above sea level. If the atmosphere 4600 m above sea level has a pressure of 70 kPa, what will be the volume of the balloon when it reaches sea level? Show your calculations. (Hint: atmospheric pressure at sea level is 101.3 kPa.)

Thank You for reading my post.

2. Nov 25, 2003

### Ambitwistor

For the first one: in order for the balloon to float, the weight of the helium in the balloon has to equal the weight of the air displaced by the volume of the balloon (Archimedes's principle).

For the second one: volume is inversely proportional to pressure (Boyle's law). (This is assuming the temperature remains constant.)

Last edited: Nov 25, 2003
3. Nov 25, 2003

### ShawnD

Since the baloon is not moving, the density and pressure are the same. It's very very simple. You won't need the ideal gas law for it.

Look at the gas law equation
PV = nRT
In this case, the pressure changes, the volume changes, n (moles) is the same, R is a constant, and T is the same. In this case, we can just say PV is a constant then.
PV before = PV after

1. v = 0.775m^3

2. V = 0.691L

Last edited: Nov 25, 2003
4. Nov 25, 2003

### Ambitwistor

Re: Re: 2 questions

ShawnD, I don't think you should be giving away the whole answer to anyone on the homework help forum (especially when the original poster specifically asked people not to give answers). The point is to give hints to get them thinking in the right direction, so they can solve it themselves, instead of solving it for them.

5. Nov 25, 2003

### ShawnD

Sorry, I didn't see that in his post. I just look at questions then see if I can solve it. I'll be more careful next time.

6. Nov 26, 2003

### Security

So what kind of equation would have to be used in the first question? Im not good at using calculations.

7. Nov 26, 2003

### Security

I should mention that the teachers' notes says that for question 1 that im suppose to look for volume = remember Density = mass/volume
An the second questions hints to consider P1V1 = P2V2

8. Nov 26, 2003

### ShawnD

For the first question, it looks like this:
density = density
1 / x = 1.29 / 1
x = 1 / 1.29
x = 0.775m^3

Second one:
PV = PV
101.3x = (1)(70)
x = 70 / 101.3
x = 0.691L

Do you see how I got those?

9. Nov 26, 2003

### Security

whats the X for?

10. Nov 26, 2003

### chroot

Staff Emeritus
ShawnD is using 'x' to represent the unknown quantity. In both problems, he chose to let x stand for the unknown volume.

Do you understand how he set up those equations? If not, let me know, and I'll help you understand them.

- Warren

11. Nov 26, 2003

### Security

Where did the 4600 m go to in the second equation?

12. Nov 26, 2003

### chroot

Staff Emeritus
It is extraneous. The problem provided the initial pressure (70 kPa) and the final pressure (101.3 kPa), which are all you need to figure the change in volume. In other words, it doesn't matter at what altitude the pressure is 70 kPa.

You might want to consider that the 4600 m and the 70 kPa are basically representing the same information. You could calculate one from the other, assuming you knew the equation that relates them.

- Warren

13. Nov 26, 2003

### Security

so does L in '0.691L' mean volume? volume = L

14. Nov 26, 2003

### chroot

Staff Emeritus
L is an abbreviation for 'liter,' which is a unit of volume. You're probably familiar with a two-liter (2L) soda bottle?

- Warren

15. Nov 26, 2003

### Security

This isnt a hard question, but its puzzling. On a question that reads: "Which of the following is true of both liquids and gases? (Choose as many as apply.)" So I choose two things that seem to be the answers: A.They take the shape of their containers - and - B.They exert a buoyant force on a submerged object equal to the weight of the liquid of gas which has been displaced

So there werent any marks from the teacher that would leave me to believe that these answers were wrong. But then she put in red ink next to my answers 'and__,___' which says to me that the remaining answers are right also. Its puzzling because I dont know if by choose as many as apply means that theirs a possiblity that all the mutliple answers are correct. Here are the remaining questions:

C.When the speed of liquid or gas moving in a ahorizontal direction increases, its pressure decreases.
D.When at rest in an enclosed system, any increase or decrease in pressure is transmitted throughout the liquid or gas.

16. Nov 26, 2003

### chroot

Staff Emeritus
All of those (A, B, C, and D) are correct.

- Warren

17. Nov 26, 2003

### Security

When submerged in water, a 2 liter object which weighs 60 N in air is buoyed up by a force of 20 N. What is the apparent weight of the object? I said 40N which is correct.

The next question asks What is the weight of the water displaced by the object in the previous question, I put down mistakenly two answers which are 80 N and 60 N. Teacher says its either one or the other. I imagine that the when you add the 60 N to the 20 N displaced that it would come to 80. But the underwater 40 N plus the water displaced comes to 60 N. I dont know which weigh to use.

18. Nov 26, 2003

### ShawnD

Ok so the object is 60N in air, the water pushes up 20N.

Since the water pushed up with a force of 20N, the weight of the water displaced by the object is 20N