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DENIUS [597]
2 years ago
14

The air pressure in Denver, Colorado is about 83 percent of the air pressure in Miami Beach, which is at about sea level. Given

that the air pressure in Miami Beach is 760 mm Hg and oxygen makes up 21 percent of the air in Miami Beach, what are the percentage of oxygen and the partial pressure of oxygen in Denver
Chemistry
1 answer:
Pani-rosa [81]2 years ago
8 0

Answer:

Explanation:

mass ratio of oxygen and nitrogen in air at Miami

= 21 : 79

ratio of their moles

= \frac{21}{32} : \frac{79}{28}     ( mol weight of oxygen is 32 and of nitrogen is 28 )

=  .65625 : 2.8214

= 1 : 4.3

This ratio will also be maintained in the air of Denver though total pressure decreases there.

Partial pressure of oxygen in air at both the places

mole fraction of oxygen

= \frac{1}{( 1 +4.3)}

= .18868

partial pressure of oxygen at Denver

= .18868 x .83 x 760

= 119 mm.

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Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

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c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

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                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

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Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

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                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

                \Delta H = n C_p \Delta T

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

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Answer:

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Answer:

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slamgirl [31]

Answer:

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Explanation:

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