Respuesta :
Answer:
Explanation:
If P₁(p) be the partial pressure of gas in first bulb then
P₁(p) (1 +1.5) = P₁V₁ = 1 X 0.75
P₁(p) = (1 / 2.5) x .75 = 0.3 atm
Similarly partial pressure of gas in second bulb
= ( 1.5 / 2.5) x 1.2 = .72 atm
Total pressure= .3 + .72 = 1.02 atms.
If n₁ and n₂ be their moles in the respective bulbs
P₁V₁ = n₁ R t
P₂ V₂ = n₂ R t
[tex]\frac{P_1V_1}{P_2V_2} = \frac{n_1}{n_2}[/tex]
[tex]\frac{n_1}{n_2+n_1 } = \frac{P_1V_1}{P_1V_1+P_2V_2}[/tex]
mole fraction of argon = [tex]\frac{.75\times 1}{.75\times1+1.2\times1.5}[/tex]
=[tex]\frac{.75}{2.55}[/tex]
= 15 / 51
Mole fraction of helium = 1 - 15 / 51 = 36 / 51
Answer:
The total pressure of the mixture is 1.02 atm.
Partial pressure of the argon : [tex]p_{Ar}=0.30atm[/tex]
Partial pressure of the helium : [tex]p_{He}=0.72atm[/tex]
Explanation:
Volume of argon in the bulb = [tex]V_1[/tex] = 1.00 L
Pressure of argon in the bulb = [tex]P_1[/tex] = 0.75 atm
Temperature of the both the gases = T
Moles of argon gases = [tex]n_1[/tex]
[tex]P_1V_1=n_1RT[/tex]..[1]
Volume of helium in the bulb = [tex]V_2[/tex] = 1.50 L
Pressure of helium in the bulb = [tex]P_2[/tex] = 1.20 atm
Moles of helium gases = [tex]n_2[/tex]
[tex]P_2V_2=n_2RT[/tex]..[2]
[1] ÷ [2]
[tex]\frac{P_1V_1}{P_2V_2}=\frac{n_1}{n_2}[/tex]
[tex]\frac{n_1}{n_2}=\frac{0.75 atm\times 1.00 L}{1.20 atm\times 1.50 L}[/tex]
[tex]\frac{n_1}{n_2}=\frac{5}{12}[/tex]
[tex]n_2=2.4\times n_1[/tex]
After opening the stopcock, the gases are mixed.
The mole fraction of argon =[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{n_1}{n_1+2.4n_1}=0.2941[/tex]
The mole fraction of helium=[tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{2.4n_1}{n_1+2.4n_1}=0.7059[/tex]
Volume of the mixture ,V= 1.00 L + 1.50 L =2.50 L
Total pressure in the mixture = P
Temperature is same = T
Total moles of gases in the mixtures = n
[tex]PV=nRT[/tex]
[tex]n=n_1+n_2[/tex]
[tex]\frac{PV}{RT}=\frac{P_1V_1}{RT}+\frac{P_1V_1}{RT}[/tex]
[tex]P\times 2.50L=0.75 atm\times 1.00 L+1.20 atm\times 1.50 L[/tex]
[tex]P\times 2.50L=2.55 atm L[/tex]
[tex]P=\frac{2.55 atm L}{2.50 L}=1.02 atm[/tex]
The total pressure of the mixture is 1.02 atm.
Partial pressure of the argon after mixing : [tex]p_{Ar}[/tex]
[tex]p_{Ar}=P\times \chi_1=1.02\times 0.2941=0.30 atm[/tex]
Partial pressure of the helium after mixing : [tex]p_{He}[/tex]
[tex]p_{He}=P\times \chi_2=1.02\times 0.7059=0.72 atm[/tex]
