Question

A mixture of gases contains 4.46 moles of Neon, 0.74 moles of Argon and 2.15
moles of Xenon. Calculate the partial pressure of gases if the total pressure is 2 atm
at certain temperature..

Answer 1

A mixture of gases contains 4.46 moles of Neon, 0.74 moles of Argon and 2.15 moles of Xenon.

Given that, nNe (number of moles of Neon) = 4.46 moles

nAr (number of moles of Argon) = 0.74 moles

nXe (number of moles of Xeon) = 2.15 moles

We have to calculate the partial pressure of gases if the total pressure is 2 atmosphere at a certain temperature.

Also given that, Total pressure = 2 atmosphere

Now,

P(Neon) = X(Neon) × P(Total pressure)

X(Ne) = (nNe)/(nNe + nAr + nXe)

X(Ne) = 4.46/(4.46 + 0.74 + 2.15)

X(Ne) = 4.46/7.35

X(Ne) = 0.607

P(Neon or Ne) = 0.607 × 2

P(Ne) = 1.21 atm

Similarly, for Argon

X(Ar) = (nAr)/(nNe + nAr + nXe)

X(Ar) = 0.74/(4.46 + 0.74 + 2.15)

X(Ar) = 0.10

P(Ar) = 0.10 × 2

P(Ar) = 0.20 atm

For Xenon

X(Xe) = (nXe)/(nNe + nAr + nXe)

X(Xe) = 2.15/(4.46 + 0.74 + 2.15)

X(Xe) = 0.293

P(Xe) = 0.293 × 2

P(Xe) = 0.586 atm

Therefore,

Partial pressure = P(Ne) + P(Ar) + P(Xe)

= 1.21 + 0.20 + 0.586 = 1.996 atm

Answer 2

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Question :

A mixture of gases contains 4.46 moles of Neon, 0.74 moles of Argon and 2.15 moles of Xenon. Calculate the partial pressure of gases if the total pressure is 2 atm at certain temperature.

$</p><p>\tt{\orange {Step-By-Step-Explaination \: : }}$

Concept Used :

Dalton's Law Of Partial Pressure :

Partial Pressures :

$\tt{ \purple{ \implies{ \rho _ {Xe} = 0.293 \times 2 \: atm \: = 0.586 \: atm.}}}$

$\tt{ \red{ \implies{ \rho _{Ar} = 0.1 \times 2 \: atm \: = 0.20 \: atm.}}}$

$\tt{ \orange{ \implies{ \rho _ {Ne} = 0.607 \times 2 \: atm \: = 1.21\: atm.}}}$

$\tt{\green{Partial \: Pressure \: = \rho _ {Xe} + \rho _ {Ar}+ \rho _ {Ne} =1.996 \: atm. }}$