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NikAS [45]
3 years ago
12

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.​
Chemistry
2 answers:
makkiz [27]3 years ago
6 0

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

Flauer [41]3 years ago
5 0

(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;

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

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

PV = nRT

where;

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

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

Learn more here:brainly.com/question/13248885

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Answer:

#1

The temperature that this happens is called the freezing point and is the same temperature as the melting point. As more energy is put into the system, the water heats up, the molecules begin moving faster and faster until there is finally enough energy in the system to totally overcome the attractive forces.

Explanation:

#2

Heating a liquid increases the speed of the molecules. An increase in the speed of the molecules competes with the attraction between molecules and causes molecules to move a little further apart. ... A decrease in the speed of the molecules allows the attractions between molecules to bring them a little closer together.

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.
VladimirAG [237]

<u>Answer:</u>

<u>For 1:</u> The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

<u>For 2:</u> The equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

<u>For 3:</u> The equilibrium pressure of oxygen gas is 0.0577 atm

<u>Explanation:</u>

  • <u>For 1:</u>

The equation used to calculate standard Gibbs free energy change of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

For the given chemical reaction:

M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)

The equation for the standard Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]

We are given:

\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

  • <u>For 2:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol    (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

K_{eq} = equilibrium constant = ?

Putting values in above equation, we get:

10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}

Hence, the equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

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The expression of K_{eq} for above equation follows:

K_{eq}=p_{O_2}^{3/2}

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

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