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2 years ago

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A gas mixture has a total pressure of 0.51 atm and consists of He and Ne. If the partial pressure of He in the mixture is 0.32 a

tm, what is the partial pressure of the Ne in the mixture? Enter your answer in the provided box. atm

1 answer:

sp2606 [1]2 years ago

8 0

Explanation:

Relation between total pressure and mole fraction is as follows.

$P_{1} = P^{o}x_{1}$

where, $P_{1}$ = partial pressure of gas 1

$x_{1}$ = mole fraction of gas 1

$P^{o}$ = total pressure of the gases

Also, it is known that $x_{1} + x_{2}$ = 1

or, $x_{2} = 1 - x_{1}$

It is given that total pressure of gas mixture is 0.51 atm and partial pressure of helium is 0.32 atm.

Hence, calculate the mole fraction of helium gas present in the given mixture as follows.

$P_{He} = P^{o}x_{He}$

0.32 atm = $0.51 atm \times x_{He}$

$x_{He}$ = 0.63

As, $x_{He} + x_{Ne}$ = 1

so, $x_{Ne} = 1 - x_{He}$

= 1 - 0.63

= 0.37

Hence, calculate the partial pressure of Ne as follows.

$P_{total} = P_{He}x_{He} + P_{Ne}x_{Ne}$

$0.51 atm = 0.32 atm \times 0.63 + P_{Ne}\times 0.37$

$0.51 atm = 0.2016 atm + 0.37 \times P_{Ne}$

0.3084 atm = $0.37 \times P_{Ne}$

$P_{Ne} = \frac{0.3084 atm}{0.37}$

= 0.834 atm

Thus, we can conclude that the partial pressure of the Ne in the mixture is 0.834 atm.

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