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Sagot :
The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981
Computation of partial pressure and number of moles:
Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761
As we know,
[tex]\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239[/tex]
According to Dalton's Law of partial pressure-
[tex]P_i=\chi_i\times P_{total}[/tex]
Where,
[tex]P_i=[/tex]The pressure of the gas component in the mixture
[tex]\chi_i=[/tex] Mole fraction of that gas component
[tex]P_t=[/tex] The total pressure of the mixture
[tex]P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm[/tex]
Calculation:
To calculate the number of moles,
PV=nRT
[tex]n=\frac{PV}{RT}[/tex]
[tex]n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole[/tex]
Learn more about Dalton's Law of partial pressure here:
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