Physics, 23.08.2019 01:30 brittanyelliott028

A1.00-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.

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Answer from: nayelimoormann

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

$\frac{P_1V_1}{P_2V_2} = \frac{n_1}{n_2}$

$\frac{n_1}{n_2+n_1 } = \frac{P_1V_1}{P_1V_1+P_2V_2}$

mole fraction of argon = $\frac{.75\times 1}{.75\times1+1.2\times1.5}$

= $\frac{.75}{2.55}$

= 15 / 51

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

Answer from: Brittpaulina

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]

$\frac{P_1V_1}{P_2V_2}=\frac{n_1}{n_2}$

$\frac{n_1}{n_2}=\frac{0.75 atm\times 1.00 L}{1.20 atm\times 1.50 L}$

$\frac{n_1}{n_2}=\frac{5}{12}$

$n_2=2.4\times n_1$

After opening the stopcock, the gases are mixed.

The mole fraction of argon = $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{n_1}{n_1+2.4n_1}=0.2941$

The mole fraction of helium= $\chi_2=\frac{n_2}{n_1+n_2}$

$\chi_2=\frac{2.4n_1}{n_1+2.4n_1}=0.7059$

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

$\frac{PV}{RT}=\frac{P_1V_1}{RT}+\frac{P_1V_1}{RT}$

$P\times 2.50L=0.75 atm\times 1.00 L+1.20 atm\times 1.50 L$

$P\times 2.50L=2.55 atm L$

$P=\frac{2.55 atm L}{2.50 L}=1.02 atm$

The total pressure of the mixture is 1.02 atm.

Partial pressure of the argon after mixing : $p_{Ar}$

$p_{Ar}=P\times \chi_1=1.02\times 0.2941=0.30 atm$

Partial pressure of the helium after mixing : $p_{He}$

$p_{He}=P\times \chi_2=1.02\times 0.7059=0.72 atm$

Answer from: Quest

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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A1.00-l bulb and a 1.50-l bulb, connected by a stopcock, are filled, respectively, with argon at 0.7...

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