2) A hot air balloon is filled with 2,500 m^3 of dry air heated to 99 degree Cel

Question

2) A hot air balloon is filled with 2,500 m^3 of dry air heated to 99 degree Celsius at 1.005 atm of pressure. Maintaining these conditions will keep the density of the air inside the balloon at 0.9486 kg/m^3.

a. Calculate the amount of air (in kilograms) contained inside of the balloon under the described conditions. Calculate the volume of the balloon in liters.

b. Calculate the volume occupied by the same amount (in kilograms) of dry air if the temperature decreased to 60 degrees Celsius. Indicate also the density for that sample of dry air.

c. As the balloon ascends and travels to a colder area, the temperature for the air inside of it decreases to 78 degrees Celsius, then 65 degrees Celsius and finally to 59 degrees Celsius. Calculate the effect temperature will have on volume, assuming that every other factor remains the same.

d. Calculate how the changes (described in the previous question) will affect the density of the air inside the balloon if amount (mass and moles) remains constant.

e. A hot air balloon will only ascend if the air density inside of it is lower than the air density outside, but air density (outside of the balloon) decreases with altitude. Calculate the required changes to keep the density (inside the balloon) under 90% of the density of the air outside the balloon if this changes to 1.2 kg/m^3, 0.803 kg/m^3, 0.775 kg/m^3, and 0.61 kg/m^3.

Explanation / Answer

a) We can stablish a 3 rule relation with the density

1m3 ----------------0,9486Kg

2500m3----------- X

X= 2500m3*0,9486Kg/1m3

X= 2371,5Kg

The relation between cubic meter and liters is: 1m3= 1000L

1m3------------- 1000L

2500m3------X

X= 2500000L=2,5*10^6L

b)From the part a we can calculate the amount of moles with the equation of the ideal gas

PV=nRT

1005atm*2,6*10^6L=n*0,082Latm/Kmol*372K

n= 8,6*10^7mol

Now they say that having the same Kg so the mol doesn't change, so we can now calculate the volume with a diferente temperature

PV=nRT

1005atm*V= 8,6*10^7mol*0,082Latm/Kmol* 333K

V= 2,33*10^6L

density= g/ml =kg/L

Because they are telling us that we have the same mass in Kg

2371,5Kg ---------2,33*10^6L

X------------------1L

X= 0,00102Kg/L= 1,02Kg/m3

c) We need to calculate the volume to the different temperatures they are giving us so we use the same analisys than before using the ideal gas equation

1005atm*V= 8,6*10^7mol*0,082Latm/Kmol* 351K V= 2,46*10^5L Temperaturre= 273 + 780C= 351K

1005atm*V= 8,6*10^7mol*0,082Latm/Kmol* 338K V=2,37*10^5L

1005atm*V= 8,6*10^7mol*0,082Latm/Kmol* 332K V= 2,33*10^5L

We can see that when the temperature decreases the volume also decreases.

d) The density is the relation between mass and volume , if the mass stays the same and the volume decreases, the density has to increase.

e) We can say that

1,2Kg-----1m3

and we can say that 1,2Kg -------100%

Now we want to see what is the weight for a 90%

1,2Kg -------100%

X ------------- 90%

X= 1,2Kg *100%/90%= 1,08Kg

and beacuse we already said that 1,2Kg was in 1m3 and this was the 100%, then we can say that there is 1,08Kg in 1m3 that is the 90 %.

Now for 0,803Kg-----1m3------100%

X------------ 1m3-----90%

X= 0,803Kg*90%/100%= 0,72Kg/m3

0,775Kg------1m3-----100%

X---------------1m3-----90%

X= 0,70Kg/m3

0,61Kg ----1m3-----100%

X -----------1m3-----90%

X=0,55Kg/m3

This are the density we have to mantein inside the ballon to keep ascending.

From the previus questions, we can say that if the Temperature rise the volume will rise and if the volume rises the density will lower down. Tis proporttional to Volume, and Volume is inversally proportional to density.

So for the density to go down so we can ascend in the ballon we have to make the temperature higher and higher each time.

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