Question

partial pressure

1. a gas mixture contains each of the following gases at the indicated partial pressures: n₂, 111 torr; o₂, 213 torr; and he, 102 torr. what is the total pressure of the mixture? what mass of each gas is present in a 1.55-l sample of this mixture at 25.0 °c?

missed this? read section 6.6; watch kcv 6.6, iwe 6.9

Explanation:

Step1: Calculate total pressure

To find the total pressure ($P_{total}$) of a gas mixture, we use Dalton's Law of Partial Pressures, which states that the total pressure is the sum of the partial pressures of the individual gases. The partial pressures are $P_{N_2} = 111$ torr, $P_{O_2} = 213$ torr, and $P_{He} = 102$ torr. So, $P_{total}=P_{N_2}+P_{O_2}+P_{He}$.
$$P_{total}=111 + 213+ 102 = 426\text{ torr}$$

Step2: Convert total pressure to atm (for ideal gas law)

We know that 1 atm = 760 torr. So, $P_{total}=\frac{426}{760}\text{ atm}\approx0.5605\text{ atm}$. The volume $V = 1.55$ L, temperature $T = 25.0^\circ\text{C}=25 + 273.15 = 298.15$ K, and $R = 0.0821\frac{\text{L·atm}}{\text{mol·K}}$.

Step3: Find total moles using ideal gas law

The ideal gas law is $PV = nRT$. Solving for $n$ (total moles), we get $n=\frac{PV}{RT}$.
$$n=\frac{0.5605\times1.55}{0.0821\times298.15}\approx\frac{0.8688}{24.48}\approx0.0355\text{ mol}$$

Step4: Find mole fraction of each gas

Mole fraction of $N_2$: $X_{N_2}=\frac{P_{N_2}}{P_{total}}=\frac{111}{426}\approx0.2606$
Mole fraction of $O_2$: $X_{O_2}=\frac{213}{426}=0.5$
Mole fraction of $He$: $X_{He}=\frac{102}{426}\approx0.2394$

Step5: Find moles of each gas

Moles of $N_2$: $n_{N_2}=X_{N_2}\times n_{total}=0.2606\times0.0355\approx0.00925\text{ mol}$
Moles of $O_2$: $n_{O_2}=X_{O_2}\times n_{total}=0.5\times0.0355 = 0.01775\text{ mol}$
Moles of $He$: $n_{He}=X_{He}\times n_{total}=0.2394\times0.0355\approx0.0085\text{ mol}$

Step6: Calculate mass of each gas

• Mass of $N_2$: Molar mass of $N_2 = 28.02$ g/mol. $m_{N_2}=n_{N_2}\times28.02=0.00925\times28.02\approx0.259\text{ g}$
• Mass of $O_2$: Molar mass of $O_2 = 32.00$ g/mol. $m_{O_2}=n_{O_2}\times32.00=0.01775\times32.00 = 0.568\text{ g}$
• Mass of $He$: Molar mass of $He = 4.003$ g/mol. $m_{He}=n_{He}\times4.003=0.0085\times4.003\approx0.0340\text{ g}$

Answer:

Total pressure: $\boldsymbol{426}$ torr
Mass of $N_2$: $\boldsymbol{\approx0.259}$ g
Mass of $O_2$: $\boldsymbol{\approx0.568}$ g
Mass of $He$: $\boldsymbol{\approx0.0340}$ g