PHYSICS HIGH SCHOOL

A 1.00-L bulb and a 1.50-L bulb, connected by a stopcock, are filled, respectively, with argon at 0.75 atm and helium at 1.20 atm at the same temperature. Calculate the total pressure and the partial pressures of each gas after the stopcock has been opened and the mole fraction of each gas. Assume ideal-gas behavior.

Answers

Answer 1

Answer:

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

$(P_1V_1)/(P_2V_2) = (n_1)/(n_2)$

$(n_1)/(n_2+n_1 ) = (P_1V_1)/(P_1V_1+P_2V_2)$

mole fraction of argon = $(.75* 1)/(.75*1+1.2*1.5)$

= $(.75)/(2.55)$

= 15 / 51

Mole fraction of helium = 1 - 15 / 51 = 36 / 51

Answer 2

Answer:

Answer:

The total pressure of the mixture is 1.02 atm.

Partial pressure of the argon : $p_(Ar)=0.30atm$

Partial pressure of the helium : $p_(He)=0.72atm$

Explanation:

Volume of argon in the bulb = $V_1$ = 1.00 L

Pressure of argon in the bulb = $P_1$ = 0.75 atm

Temperature of the both the gases = T

Moles of argon gases = $n_1$

$P_1V_1=n_1RT$ ..[1]

Volume of helium in the bulb = $V_2$ = 1.50 L

Pressure of helium in the bulb = $P_2$ = 1.20 atm

Moles of helium gases = $n_2$

$P_2V_2=n_2RT$ ..[2]

[1] ÷ [2]

$(P_1V_1)/(P_2V_2)=(n_1)/(n_2)$

$(n_1)/(n_2)=(0.75 atm* 1.00 L)/(1.20 atm* 1.50 L)$

$(n_1)/(n_2)=(5)/(12)$

$n_2=2.4* n_1$

After opening the stopcock, the gases are mixed.

The mole fraction of argon = $\chi_1=(n_1)/(n_1+n_2)$

$\chi_1=(n_1)/(n_1+2.4n_1)=0.2941$

The mole fraction of helium= $\chi_2=(n_2)/(n_1+n_2)$

$\chi_2=(2.4n_1)/(n_1+2.4n_1)=0.7059$

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

$PV=nRT$

$n=n_1+n_2$

$(PV)/(RT)=(P_1V_1)/(RT)+(P_1V_1)/(RT)$

$P* 2.50L=0.75 atm* 1.00 L+1.20 atm* 1.50 L$

$P* 2.50L=2.55 atm L$

$P=(2.55 atm L)/(2.50 L)=1.02 atm$

The total pressure of the mixture is 1.02 atm.

Partial pressure of the argon after mixing : $p_(Ar)$

$p_(Ar)=P* \chi_1=1.02* 0.2941=0.30 atm$

Partial pressure of the helium after mixing : $p_(He)$

$p_(He)=P* \chi_2=1.02* 0.7059=0.72 atm$

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