Solved Problem on Gases
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A gas is initially at a pressure of 2 atmospheres, volume of 4 liters, and temperature of 300 kelvins. From this state, the pressure increases at a constant volume up to a pressure of 5 atmospheres, determine its temperature at the end of this process. After this first process, the gas expanded at a constant temperature to a volume of 8 liters, determine its pressure at the end of this second process. From this state, the gas is heated at a constant pressure to a temperature of 900 kelvins, determine the final volume of the gas. Plot the gas processes in a graph of the pressure as a function of volume p×V.


Solution

For the initial and final states of the first process

Initial state Final state
pressure:    p0 = 2 atm pressure:    p1 = 5 atm
volume:    V0 = 4 \( \ell \) volume:    V1 = 4 \( \ell \)
temperature:    T0 = 300 K temperature:    T1

The General Gas Equation is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{p_{i}V_{i}}{T_{i}}=\frac{p_{f}V_{f}}{T_{f}}} \end{gather} \]
As in the process the volume remains constant Vi = Vf, we have an isometric or isovolumetric, or isochoric process, and the General Law becomes the Gay-Lussac's Law.
\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{p_{i}}{T_{i}}=\frac{p_{f}}{T_{f}}} \end{gather} \]
substituting the problem data
\[ \begin{gather} \frac{2}{300}=\frac{5}{T_{1}}\\[5pt] T_{1}=\frac{5\times 300}{2}\\[5pt] T_{1}=5\times 150 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T_{1}=750\;\text{K}} \end{gather} \]
After the first process, the final state of the gas becomes the initial state of the second process. For the initial and final states of the second process

Initial state Final state
pressure:    p1 = 5 atm pressure:    p2
volume:    V1 = 4 \( \ell \) volume:    V2 = 8 \( \ell \)
temperature:    T1 = 750 K temperature:    T2 = 750 K

As in the process the temperature remains constant Ti = Tf, we have an isothermal process, and the General Law becomes Boyle's Law.
\[ \begin{gather} \bbox[#99CCFF,10px] {p_{i}V_{i}=p_{f}V_{f}} \end{gather} \]
substituting the problem data
\[ \begin{gather} 5\times 4=8 p_{2}\\[5pt] p_{2}=\frac{20}{8} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {p_{2}=2.5\;\text{atm}} \end{gather} \]
After the second process, the final state of the gas becomes the initial state of the third process. For the initial and final states of the third process

Initial state Final state
pressure:    p2 = 2.5 atm pressure:    p3 = 2.5 atm
volume:    V2 = 8 \( \ell \) volume:    V3
temperature:    T2 = 750 K temperature:    T3 = 900 K

As in the process the pressure remains constant pi = pf, we have an isobaric process, and the General Law becomes Charles' Law.
\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{V_{i}}{T_{i}}=\frac{V_{f}}{T_{f}}} \end{gather} \]
substituting the problem data
\[ \begin{gather} \frac{8}{750}=\frac{V_{3}}{900}\\[5pt] V_{3}=\frac{8\times 900}{750} \end{gather} \]
dividing the numerator and the denominator for 50
\[ \begin{gather} V_{3}=\frac{8\times 900:50}{750:50}\\[5pt] V_{3}=\frac{8\times 18}{15}\\[5pt] V_{3}=\frac{144}{15} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V_{3}=9.6\;\text{l}} \end{gather} \]
With the datum obtained in the problem, we plot Graph 1. From point A to B, the gas undergoes an isometric or isovolumetric, or isochoric, process, from point B to C is an isothermal process, and from point C to D is an isobaric process.

Graph 1
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .