A gas tank contains 2.5 moles of oxygen, 7.0 moles of nitrogen and 0.5 mole of water vapor. The total pressure of the mixture is
1 answer:
Answer:
The partial pressure of each gas is:
- Oxygen= 213 torr
- Nitrogen= 596.4 torr
- Water Vapor= 42.6 torr
Explanation:
The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:
PT = PA + PB
This relationship is due to the assumption that there are no attractive forces between the gases.
Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.
The mole fraction of a gas in a gas mixture is given by:
XA = nA / nT
This fraction will always be less than 1.
So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:
PA = XA * PT
In this case, you know:
- number of moles of oxygen= 2.5 moles
- number of moles of nitrogen= 7.0 moles
- number of moles of water vapor= 0.5 moles
So, the number of total moles nT is calculated as:
nT= number of moles of oxygen + number of moles of nitrogen +number of moles of water vapor
Then:
nT= 2.5 moles + 7.0 moles + 0.5 moles
nT= 10
So, the mole fraction of the gas in the mixture is:
- mole fraction of oxygen =
= 0.25
- mole fraction of nitrogen =
= 0.7
- mole fraction of water vapor =
= 0.05
Being the total presurre (PT) equal to 852 torr, then:
- Poxygen= 0.25*852 torr= 213 torr
- Pnitrogen= 0.7*852 torr= 596.4 torr
- Pwater vapor= 0.05*852 torr= 42.6 torr
<u><em>The partial pressure of each gas is: </em></u>
- <u><em>Oxygen= 213 torr</em></u>
- <u><em>Nitrogen= 596.4 torr</em></u>
- <u><em>Water Vapor= 42.6 torr</em></u>
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Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg = = 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
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= 21.6 kj. kg⁻¹
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Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT =
⇒ ΔT =
⇒ ΔT = 5.16 K
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