Respuesta :
Answer:
the mole fraction of Gas B is xB= 0.612 (61.2%)
Explanation:
Assuming ideal gas behaviour of A and B, then
pA*V=nA*R*T
pB*V=nB*R*T
where
V= volume = 10 L
T= temperature= 25°C= 298 K
pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm
R= ideal gas constant = 0.082 atm*L/(mol*K)
therefore
nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole
nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole
therefore the total number of moles is
n = nA +nB= 2.04 mole + 3.22 mole = 5.26 mole
the mole fraction of Gas B is then
xB= nB/n= 3.22 mole/5.26 mole = 0.612
xB= 0.612
Note
another way to obtain it is through Dalton's law
P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612
The mole fraction of gas B in the 10 L vessel is 0.612
To solve this question, we'll begin by calculating the total pressure of the gas. This can be obtained as follow:
Partial pressure of Gas A = 5 atm
Partial pressure of Gas B = 7.89 atm
Total pressure =?
Total pressure = Partial pressure of Gas + Partial pressure of Gas B
Total pressure = 5 + 7.89
Total pressure = 12.89 atm
Finally, we shall determine the mole fraction of Gas B. This can be obtained as follow:
Partial pressure of Gas B ([tex]P_{B}[/tex]) = 7.89 atm
Total pressure ([tex]P_{T}\\[/tex]) = 12.89 atm
Mole fraction of Gas B ([tex]n_{B}[/tex]) =?
[tex]n_{B} = \frac{P_{B} }{P_{T} } \\\\n_{B} = \frac{7.89}{12.89}\\\\n_{B} = 0.612[/tex]
Therefore, the mole fraction of Gas B in the 10 L vessel is is 0.612
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