Solved Problem on Gases

A gas initially occupies a volume of 15 liters at 27 °C and exerts a pressure of 5 atmospheres on the walls of the container where it is closed. Determine:
a) The volume occupied by the gas at 127 °C under the pressure of 5 atmospheres;
b) From the initial state, the pressure exerted by the gas at 27 °C when its volume increases to 30 liters;
c) From the initial state, the temperature of the gas, under a pressure of 8 atmospheres and 15 liter volume.

Solution

a) First, we must convert the given temperatures in degrees Celsius (°C) to Kelvins (K) used in the International System of Units (S.I.)

\[ \begin{gather} T_{i}=t_{i}+273=27+273=300\;\text{K}\\[10pt] T_{f}=t_{f}+273=127+273=400\;\text{K} \end{gather} \]

For the initial and final states

Initial state	Final state
pressure: p_i = 5 atm	pressure: p₁ = 5 atm
volume: V_i = 15 \( \ell \)	volume: V₁
temperature: T_i = 300 K	temperature: T₁ = 400 K

From the General Gas Equation

\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{p_{i}V_{i}}{T_{i}}=\frac{p_{f}V_{f}}{T_{f}}} \end{gather} \]

a) As in the given process, the pressure remains constant p_i = p_f, 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

\[ \begin{gather} \frac{15}{300}=\frac{V_{1}}{400}\\[5pt] v_{1}=\frac{15\times 400}{300}\\[5pt] v_{1}=\frac{15\times 4}{3}\\[5pt] v_{1}=5\times 4 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V_{1}=20\;\ell} \end{gather} \]

b) For the initial and final states

Initial state	Final state
pressure: p_i = 5 atm	pressure: p₂
volume: V_i = 15 \( \ell \)	volume: V₂ = 30 \( \ell \)
temperature: T_i = 300 K	temperature: T₁ = 300 K

As in the given process, the temperature remains constant T_i = T_f, 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

\[ \begin{gather} 5\times 15=30 p_{2}\\[5pt] p_{2}=\frac{75}{30} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p_{2}=2.5\;\text{atm}} \end{gather} \]

c) For the initial and final states

Initial state	Final state
pressure: p_i = 5 atm	pressure: p₃ = 8 atm
volume: V_i = 15 \( \ell \)	volume: V₂ = 15 \( \ell \)
temperature: T_i = 300 K	temperature: T₃

As in the given process, the volume remains constant V_i = V_f, we have an isovolumetric or isometric, or isochoric process, and the General Law becomes 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

\[ \begin{gather} \frac{5}{300}=\frac{8}{T_{3}}\\[5pt] T_{3}=\frac{8\times 300}{5}\\[5pt] T_{3}=8\times 60 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T_{3}=480\;\text{K}} \end{gather} \]

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