determine the total pressure of gas mixture the contains oxygen, nitrogen, and helium if the partial pressures of the gases are
1 answer:
0.922 atm is the total pressure of gas mixture that contains oxygen, nitrogen and helium gases.
Explanation:
Data given:
partial pressure of the oxygen p = 0.197 atm
partial pressure of the nitrogen gas p = 0.461 atm
partial pressure of the helium gas pHe = 0.264
total pressure P in the gas mixture= ?
We know that individual gases exert pressure on the mixture of gases independently. From Dalton's law of pressure it is said that sum of partial pressure of all the gases in the mixture equals its total pressure. So,
Ptotal = p + p
+ pHe
putting the values in the mixture:
Ptotal = 0.197 + 0.461+ 0.264
Ptotal = 0.922 atm
Total pressure is 0.922 atm.
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Answer:
You must remove
.
Explanation:
There are three heat transfers in this process:
Total heat = cool the vapour + condense the vapour + cool the liquid
q = q₁ + q₂ + q₃
q = nC₁ΔT₁ + nΔHcond + nC₂ΔT₂
Let's calculate these heat transfers separately.
Data:
You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.
C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹
C₂ = specific heat capacity of liquid = 115 J·K⁻¹mol⁻¹
ΔHcond = -38.56 kJ·mol⁻¹
Tmax = 300 °C
b.p. = 78.4 °C
Tmin = 25.0 °C
n = 0.782 mol
Calculations:
ΔT₁ = 78.4 - 300 = -221.6 K
q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ
q₂ = 0.782 × (-38.56) = -30.15 kJ
ΔT = 25.0 - 78.4 = -53.4 K
q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ
q = -15.60 - 53.4 - 4.802 = -50.6 kJ
You must remove of heat to convert the vapour to a gas.
The answer is either 38 g or 40.g KCLO3. If it 95% instead of 895%.
