Question

Hot air balloons are open at their lower and end. A basket is suspended from the balloon and can take several passengers as well as the pilot. A propane burner fixed to the basket is used to heat up the air in the balloon A hot balloon has a volume of 1500 m3 and its volume does not change when the enclosed air is heated.
1) Use the ideal gas equation, pV=nRT. to calculate how many molecules of the balloon contains at a temperature of 10 °C and a pressure 10.1 KP.
2) Calculate the mass of air it contains, assuming an average M of 29.​

Answer 1

Answer:

using pv=nRt

Explanation:

p= 10.1 kp

v= 1500m3

n=?

R= 8.314

T= 10 + 273= 283k

pv = nRt

10.1 *1500=n*8.314*283

n= 10.1*1500/8.314*283

n= 15150/2352.862

n= 6.44 moles

2) n= mass/ molar mass

6.44= mass/ 29

mass= 29* 6.44

mass= 186.76g

Answer 2

(a) The number of molecules contained by the balloon at the given condition is 6,438.97 moles.

(b) The mass of air in the balloon is 186.73 kg.

The given parameters;

• volume of the balloon, V = 1500 m³
• temperature, T = 10⁰C = 273 + 10 = 283 K
• pressure, = P = 10.1 kPa = 10,100 Pa

The number of molecules contained by the balloon at the given condition is calculated as follows;

$PV = nRT$

where;

• R is ideal gas constant = 8.314 J/mol.K
• n is the number of moles air in the balloon;

$n = \frac{PV}{RT} \\\\n = \frac{10,100 \times 1500}{8.314 \times 283} \\\\n = 6,438.97 \ moles$

The mass of air in the balloon is calculated as;

$n = \frac{m}{M} \\\\m = nM\\\\m = 6,438.97\ moles \ \times \ 29 \ g/mol\\\\m = 186.73 \ kg$

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