Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q5P

Expert-verified
Physics Principles with Applications
Found in: Page 412
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Question: (II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

The P-V diagram is shown below:

See the step by step solution

Step by Step Solution

Step 1: Understanding of isobaric process 

The isobaric process may be described as the thermodynamic process that happens at constant pressure. The volume of the thermodynamic system varies during the isobaric process.

Step 2: Given information 

Given data:

The initial volume of the air is \[{V_{\rm{i}}} = 1.0\;{\rm{L}}\].

The initial gauge pressure of the air is \[{P_{{\rm{i,g}}}} = 3.5\;{\rm{atm}}\].

The final gauge pressure of the air is \[{P_{{\rm{f,g}}}} = 1.0\;{\rm{atm}}\].

Step 3: Evaluation of the final volume of the air                         

The initial absolute pressure of the air can be calculated as:

\[\begin{array}{l}{P_{{\rm{i,a}}}} = {P_{{\rm{i,g}}}} + \left( {1.0\;{\rm{atm}}} \right)\\{P_{{\rm{i,a}}}} = \left( {3.5\;{\rm{atm}}} \right) + \left( {1.0\;{\rm{atm}}} \right)\\{P_{{\rm{i,a}}}} = 4.5\;{\rm{atm}}\end{array}\]

The final absolute pressure of the air can be calculated as:

\[\begin{array}{l}{P_{{\rm{f,a}}}} = {P_{{\rm{f,g}}}} + \left( {1.0\;{\rm{atm}}} \right)\\{P_{{\rm{f,a}}}} = \left( {1.0\;{\rm{atm}}} \right) + \left( {1.0\;{\rm{atm}}} \right)\\{P_{{\rm{f,a}}}} = 2.0\;{\rm{atm}}\end{array}\]

From ideal gas equation, the relation between the pressure and volume is given as:

\[PV = {\rm{constant}}\]

For any two states, the above equation can be written as:

\[\begin{array}{c}{P_{{\rm{i,a}}}}{V_{\rm{i}}} = {P_{{\rm{f,a}}}}{V_{\rm{f}}}\\{V_{\rm{f}}} = \frac{{{P_{{\rm{i,a}}}}{V_{\rm{i}}}}}{{{P_{{\rm{f,a}}}}}}\end{array}\]

Substitute the values in the above equation.

\[\begin{array}{l}{V_{\rm{f}}} = \frac{{\left( {4.5\;{\rm{atm}}} \right)\left( {1.0\;{\rm{L}}} \right)}}{{\left( {2.0\;{\rm{atm}}} \right)}}\\{V_{\rm{f}}} = 2.25\;{\rm{L}}\end{array}\]

Thus, the final volume of the air is \[2.25\;{\rm{L}}\].

Step 4: P-V diagram for the given values

The P-V diagram for the given values can be drawn as:

In the above P-V diagram, curve AB represents the isothermal expansion. In this process pressure changes from \(4.5\;{\rm{atm}}\) to \(2.0\;{\rm{atm}}\) and volume changes from \(1.0\;{\rm{L}}\)to \(2.25\;{\rm{L}}\). The curve BC represents the air compressed at constant absolute pressure of \(2.0\;{\rm{atm}}\). The curve CA represents the air brought back to initial absolute pressure of \(4.5\;{\rm{atm}}\) at constant volume.

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.