Partial pressure

Chemistry Level 2

In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. Consider the three flasks in the diagram above. When the stopcocks are closed, the three flasks have the following volume and pressure values: He $1.0$ L at $1$ atm, vacuum $2.0$ L, Ar $2.0$ L at $2$ atm. Assuming the connecting tubes have negligible volume, what is the partial pressure of each gas and the total pressure after all the stopcocks are opened?

He: 0.2 atm, Ar: 0.8 atm, Total: 1.0 atm He: 0.2 atm, Ar: 0.8 atm, Total: 0.5 atm He: 0.2 atm, Ar: 0.4 atm, Total: 0.6 atm He: 0.4 atm, Ar: 0.8 atm, Total: 1.2 atm

2 solutions

Chew-Seong Cheong
Feb 24, 2016

Assuming He and Ar behave like ideal gases, then at a constant temperature $pV = constant$ , where $p$ is the pressure and $V$ , volume. Therefore:

$\begin{aligned} p'_{He} V'_{He} & = p_{He} V_{He} \\ \Rightarrow p'_{He} & = \frac{p_{He} V_{He}}{V'_{He}} \\ & = \frac{1 \times 1}{5} = 0.2 \text{ atm} \end{aligned}$

Similarly, $\begin{aligned} p'_{Ar} & = \frac{2 \times 2}{5} = 0.8 \text{ atm} \end{aligned}$

Therefore, the answer is $\boxed{\text{He: 0.2 atm, Ar: 0.8 atm, total: 1 atm}}$

Lu Chee Ket
Jan 30, 2016

By ordinary way of $\Sigma$ (magnitude $\times$ quantity) = Total product:

1 $\times$ 1 + 0 $\times$ 2 + 2 $\times$ 2 = $\Large P_{resultant}$ $\times$ (1 + 2 + 2)

$\Large P_{resultant}$ = $\frac{1 + 0 + 4}{1 + 2 + 2} = 1~atm$

$Partial~P_1 = \frac{1 \times 1}{1 \times (1 + 2 + 2)} = 0.2$ and

$Partial~P_3 = \frac{2 \times 2}{1 \times (1 + 2 + 2)} = 0.8$

Note that 0.2 + 0.8 = 1.0

Answer: $\boxed{He: 0.2~atm, Ar: 0.8~atm, Total: 1.0~atm}$

Partial pressure

2 solutions

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