Question

vessel contain 0.25 moles of Helium and 0.15 moles of neon at 298K and 2.4 atmosphere pressure. Calculate the partial pressure of He and Ne respectively​

Answer 1

Explanation:

The mole fraction of carbon dioxide =

total number of moles of mixture

number of moles of carbon dioxide

=

0.5+0.1+0.4

0.1

=0.1

The partial pressure of carbon dioxide = mole fraction of carbon dioxide× total pressure =0.1×5=0.5 atm.

Answer 2

Answer:

The partial pressure of Helium, $P_{He}$, measured is $1.5atm$.

The partial pressure of Neon, $P_{Ne}$, measured is $0.9atm$.

Explanation:

Given,

The number of moles of helium, $n_{He}$ = $0.25moles$

The number of moles of neon, $n_{Ne}$ = $0.15moles$

The given total atmospheric pressure = $2.4atm$

The partial pressure of Helium, $P_{He}$ =?

The partial pressure of Neon, $P_{Ne}$ =?

As we know,

• Partial pressure = mole fraction × total atmospheric pressure

Firstly, we have to calculate the mole fractions.

• The mole fraction of Helium = $\frac{n_{He} }{n_{He} + n_{Ne} }$ = $\frac{0.25}{0.25 + 0.15}$ = $0.625$
• The mole fraction of Neon = $\frac{n_{Ne} }{n_{He} + n_{Ne} }$ = $\frac{0.15}{0.25 + 0.15}$ = $0.375$

Now,

• The partial pressure of Helium, $P_{He}$ = mole fraction of Helium × total atmospheric pressure
• The partial pressure of Helium, $P_{He}$ = $0.625 \times 2.4$ = $1.5atm$.

Also,

• The partial pressure of Neon, $P_{Ne}$ = mole fraction of Neon × total atmospheric pressure
• The partial pressure of Neon, $P_{Ne}$ = $0.375\times 2.4$ = $0.9atm$.