Question

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa, then what is the partial pressure of helium?

Answer 1

Answer : The partial pressure of helium is, $1.815\times 10^4KPa$

Solution : Given,

Molar mass of $O_2$ = 32 g/mole

Molar mass of helium = 4 g/mole

Molar mass of $N_2$ = 28 g/mole

Total pressure of gas = $2.07\times 10^4KPa$

As we are given gases in percent, that means 10 g of oxygen gas, 50 g of helium gas and 40 g of nitrogen gas present in 100 g of mixture.

First we have to calculate the moles of oxygen, helium and nitrogen gas.

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{10g}{32g/mole}=0.3125moles$

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{50g}{4g/mole}=12.5moles$

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{40g}{28g/mole}=1.428moles$

Now we have to calculate the total number of moles of gas mixture.

$\text{Total number of moles of gas}=\text{Moles of oxygen gas}+\text{Mole of helium gas}+\text{Moles of nitrogen gas}$

$\text{Total number of moles of gas}=0.3125+12.5+1.428=14.24moles$

Now we have to calculate the moles fraction of helium gas.

$\text{Mole fraction of He gas}=\frac{\text{Moles of He gas}}{\text{Total number of moles of gas}}=\frac{12.5}{14.25}=0.877$

Now we have to calculate the partial pressure of helium.

$p_{He}=X_{He}\times P_T$

where,

$p_{He}$ = partial pressure of helium

$P_T$ = total pressure

$X_{He}$ = mole fraction of helium

Now put all the given values in this formula, we get

$p_{He}=(0.877)\times (2.07\times 10^4KPa)=1.815\times 10^4KPa$

Therefore, the partial pressure of helium is, $1.815\times 10^4KPa$