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Strike441 [17]
3 years ago
12

If a helium balloon is put into a hot oven. What happens to the volume of the balloon?

Chemistry
1 answer:
AveGali [126]3 years ago
6 0

Answer:

What about hot air balloons? They work by similar principles. If you heat up a gas it expands. In the case of a hot air balloon, when the gas inside the balloon expands the extra gas is pushed out the bottom of the balloon, meaning that there are fewer atoms inside the balloon, meaning that the air in the balloon is lighter than the air outside the balloon.

The amount of lifting power is controlled by how hot the air is. If you heat the air inside the balloon 100 degrees F hotter than the outside air temperature, then the air inside the balloon will be about 25 percent lighter than the air outside the balloon. So a cubic foot of air weighs about 35 grams at 32 degrees F. A cubic foot of hot air at 132 degrees F will weigh 25 percent less, or about 26.5 grams. The difference is 8.5 grams or so. So a hot air balloon has to be much bigger to support the same weight, but it will float because hotter air is lighter than cooler air.

Explanation:

The amount of lifting power

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Catalyst
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a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

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