Question

when fully inflated a hot air balloon has a volume of 1.6 *10^5 L ( liter) an average temperature of 373k and 0 967 atm . 1) Assuming that the air has an average molar mass of 29g/ mol what is the density of the air in the hot air balloon? P = .... g/L . (2) how does this compare with the density of air at STP ? (a) . it is less dense than at STP . (b) it is more dense than at STP .

Answer 1

1.  First, you have to find the number of moles 1.6z10^5L of gas is at 373K and 0.967atm using PV=nRT solving for n. (n=PV/RT).  Everything is in the correct units and we know R is going to be 0.08206atmL/molK since it is a constant.  
n=(0.967atmx160000L)/(0.08206atmL/molKx373K)
n=5054.8mol gas
Then you have to find the the number grams which can be found using the molar mass given as 29g/mol.  multiply 29g/mol by the number of moles of gas we found in the previous step.
5054.8molx29g/mol=146589.9g of gas
Lastly, to find the density of the gas you need to divide the mass of the gas by its volume.
146589.9g/160000L=0.916g/L

2. The dinsity of the gas at STP should be higher than the density of gas with the given conditions.  This is due to the fact that the given conditions involves a higher temperature than that of at STP which will cause the gas to expand therefore increasing the volume with out increasing the mass.  The reason why the pressure is not building up even though the pressure is higher is that the balloon is not sealed meaning the gas can maintain about atmospheric pressure while expanding since the excess are just leaves the balloon.
the answer to part 2 can be proven by the fallowing:
To find the density of the gas at STP you first multiply the molar volume of gas at STP by the number of moles of gas from part 1 to get the volume of the gas at STP.
5054.8molx22.4L/mol=113228L
Then you divide the mass form part by the new volume to get the new density.
146589.9g/113228L=1.30g/L

I hope this helps.  Let me know in the comments if any of it is unclear. 

Answer 2

Answer:

0.91 g/L is the density of the air in the hot air balloon.

1.29 g/L at STP, which means gas is less dense than at STP.

Explanation:

Pressure of gas,P = 0.967 atm

Temperature of the gas = T = 373 K

Volume of the gas  = V = $1.6\times 10^5 L$

Mass of the gas = m

Molar mass,M = 29 g/mol

Density of air in the gas at given conditions: d

Using an ideal gas equation:

$PV=nRT=\frac{m}{M}RT$

$PM=\frac{m}{V}RT=dRT$

(Density =$\frac{Mass}{Volume}$)

$d=\frac{PM}{RT}=\frac{0.967 atm\times 29 g/mol}{0.0821 atm L/mol K\times 373 K}=0.91 g/L$

0.91 g/L is the density of the air in the hot air balloon.

Density of air in the gas at STP conditions: d'

Pressure of gas at STP = P' = 1 atm

Temperature of the gas at STP = T' = 273 K

$d'=\frac{P'M}{RT'}=\frac{1 atm\times 29 g/mol}{0.0821 atm L/mol K\times 273 K}=1.29 g/L$

d' > d

The gas is less dense than at STP