# Math Help - Two air pressure/density questions that I have little clue what to do with.

1. ## Two air pressure/density questions that I have little clue what to do with.

Hello again MHF!

I have a some questions that I need a little help with concerning air density and buoyancy.

(1):
The density of air near the earth's surface is 1.29 kg/m^3. If a helium balloon with a mass of 1kg floats in air with out rising or falling, what is the minimum volume of the helium in the balloon? Show your calculations.(presume that the mass of the material making up the balloon is negligible.)

(2):
A balloon filled with one liter of air is tied to a brick and dropped from a height of 4600m above sea level. If the atmosphere 4600m above sea level has a pressure of 70kPa, what will the volume of the balloon be when it reaches sea level? Show your calculations.

Same story sadly, I have no idea what to do with this one.

2. Originally Posted by MathBlaster47
Hello again MHF!

I have a some questions that I need a little help with concerning air density and buoyancy.

(1):
The density of air near the earth's surface is 1.29 kg/m^3. If a helium balloon with a mass of 1kg floats in air with out rising or falling, what is the minimum volume of the helium in the balloon? Show your calculations.(presume that the mass of the material making up the balloon is negligible.)

(2):
A balloon filled with one liter of air is tied to a brick and dropped from a height of 4600m above sea level. If the atmosphere 4600m above sea level has a pressure of 70kPa, what will the volume of the balloon be when it reaches sea level? Show your calculations.

Same story sadly, I have no idea what to do with this one.

To #2:

2. You are supposed to know the air pressure at sea level (at 20°C). Then

$p_1 \cdot V_1 = p_2 \cdot V_2$

with p_1 = 70 kPA, V_1 = 1 dm³, p_2 = you know that (101.3 kPa), V_2 = you calculate that!

3. Originally Posted by MathBlaster47
Hello again MHF!

I have a some questions that I need a little help with concerning air density and buoyancy.

(1):
The density of air near the earth's surface is 1.29 kg/m^3. If a helium balloon with a mass of 1kg floats in air with out rising or falling, what is the minimum volume of the helium in the balloon? Show your calculations.(presume that the mass (<--- do you mean volume?) of the material making up the balloon is negligible.)

...

1. A solid surrounded by a gas experiences a force of buoyancy(?) which has the opposite direction of the weight of the solid and has the absolute value of the weight of the displaced gas. (Principle of Archimedes)

2. Since the body looses all it's weight you get as force of buoyancy $F_b = 1 kg \cdot 9.81 \frac{m}{s^2} = 9.81 N$

3. The densitiy of the surrounding air is $d_a = 1.29 \frac{kg}{m^3}$

The force of buoyancy is calculated by: $F_b = V_{air} \cdot d_a \cdot 9.81 \frac{m}{s^2}$

4. Solve for $V_{air}$ which must be as large as the volume of the Helium gas.
Spoiler:
I've got 0.775 m³

4. Problem 1. 1 kg of helium = 1/4 kg mol STP volume =.25x 22.4 cume=5.6cume
correct only for a temp of 293 deg K 20+273 5.6x293/273=6.0 cu me

Problem 2. Molal volume of air @ the 4600m elevation 70 kPa and temp not given can be calculated by specifying temp.
Use gr mols here 29 grs 0f air = 1mol or 22.4 liters STP 1 liter is what percent of the actual volume calculated.This percent is the fraction of 1gr mol put into balloon.Calculate the volume of this weight at the conditions @ sea level

5. Originally Posted by earboth
To #2:

2. You are supposed to know the air pressure at sea level (at 20°C). Then

$p_1 \cdot V_1 = p_2 \cdot V_2$

with p_1 = 70 kPA, V_1 = 1 dm³, p_2 = you know that (101.3 kPa), V_2 = you calculate that!
Ok, I'm a little confused by "dm^3". Did you mean cm^3?
Either way, am I getting it right that all I have to do is multiply the volume(1L, I think) by the pressure(70kPa) and then divide that number by the new pressure in order to find the new volume?

6. Hi mathblaster47
Yes this is true but you assume that the temperatures are the same.For a specific volume of gas at given conditions

P1v1/T1= P2v2/T2

bjh

7. Originally Posted by bjhopper
Hi mathblaster47
Yes this is true but you assume that the temperatures are the same.For a specific volume of gas at given conditions

P1v1/T1= P2v2/T2

bjh
The question did not give me any temperatures to work with, so I don't know what to do about that.

8. Originally Posted by MathBlaster47
Ok, I'm a little confused by "dm^3". Did you mean cm^3?
...
Sorry for the confusion....

You wrote: "A balloon filled with one liter of air ...". 1 l is the volume of a cube with the side-length = 10 cm = 1 dm (<--- that's 1 deci-meter)

In Germany liter is mostly used for liquids. Therefore I used the mathematically correct unit of a volume.

9. Hello mathblaster47,

Standard values of temperature and pressure above sea level

Sea level 15 C 101.3kPa
15000 ft -15 C 57 kPa

bjh

10. Originally Posted by earboth
Sorry for the confusion....

You wrote: "A balloon filled with one liter of air ...". 1 l is the volume of a cube with the side-length = 10 cm = 1 dm (<--- that's 1 deci-meter)

In Germany liter is mostly used for liquids. Therefore I used the mathematically correct unit of a volume.
No problem, I thought that might be the case, just had to make sure!
I wish America would switch over to the metric system already.....

Originally Posted by bjhopper
Hello mathblaster47,

Standard values of temperature and pressure above sea level

Sea level 15 C 101.3kPa
15000 ft -15 C 57 kPa

bjh
The thing is, the book that I got this question from says nothing about that part of the formula, and the question doesn't ask me to take that into consideration, I will do the working for my own benefit though.
I'm just not sure if I'll get the question right or wrong if I include that in my answer.

11. Originally Posted by earboth
1. A solid surrounded by a gas experiences a force of buoyancy(?) which has the opposite direction of the weight of the solid and has the absolute value of the weight of the displaced gas. (Principle of Archimedes)

2. Since the body looses all it's weight you get as force of buoyancy $F_b = 1 kg \cdot 9.81 \frac{m}{s^2} = 9.81 N$

3. The densitiy of the surrounding air is $d_a = 1.29 \frac{kg}{m^3}$

The force of buoyancy is calculated by: $F_b = V_{air} \cdot d_a \cdot 9.81 \frac{m}{s^2}$

4. Solve for $V_{air}$ which must be as large as the volume of the Helium gas.
Spoiler:
I've got 0.775 m³
Let me make sure I have this straight:
Because the balloon has a weight of 9.81N the force of buoyancy is also 9.81N.
Since the density of the air is 1.29kg/m^3 it displaces that much air for 1kg of mass.
So, all I have to do is(ignoring the unit symbols): (9.81/9.81)/1.29, to get my answer.
Am I correct on the principle?