Solved Problem on Gases

A container is filled with a certain mass of gas at a pressure of 3.0 atm and a temperature of 27 °C. Determine the percentage of gas that must escape from the container so that its pressure at 47 °C is 2.0 atm. Neglect the expansion of the container.

Problem data:

First, we must convert temperatures, given in degrees Celsius (°C), to kelvins (K).

Initial State	Final State
p₁ = 3.0 atm	p₂ = 2.0 atm
V₁ = V	V₂ = V
T₁ = 27 °C = 300 K	T₂ = 47 °C = 320 K
n₁	n₂

Solution

Using the Equation of State for an Ideal Gas

\[ \begin{gather} \bbox[#99CCFF,10px] {pV=nRT} \end{gather} \]

we can write the initial and final states in terms of the number of moles in the container at the beginning, and what remains after letting some of it escape

\[ \begin{gather} n_{1}=\frac{p_{1}V_{1}}{RT_{1}} \end{gather} \]

\[ \begin{gather} n_{2}=\frac{p_{2}V_{2}}{RT_{2}} \end{gather} \]

where V₁ = V₂ = V and writing the ratio between the final and initial number of moles

\[ \begin{gather} \frac{n_{2}}{n_{1}}=\frac{\dfrac{p_{2}\cancel{V}}{\cancel{R}T_{2}}}{\dfrac{p_{1}\cancel{V}}{\cancel{R}T_{1}}}=\frac{p_{2}T_{1}}{p_{1}T_{2}} \end{gather} \]

substituting the given values

\[ \begin{gather} \frac{n_{2}}{n_{1}}=\frac{2.0\times 300}{3.0\times 320}\\[5pt] n_{2}=0.625n_{1} \end{gather} \]

The final number of moles will be 62.5% of the initial total. We must let escape

\[ \begin{gather} 100\text{%}-62.5\text{%}=37.5\text{%} \end{gather} \]

For the final conditions to be reached 37.5% of the gas must escape.

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