# One of the more confusing notions to beginning

• Notes
• 36

This preview shows page 15 - 18 out of 36 pages.

One of the more confusing notions to beginning students of thermodynamics is which forms of energy and specific heat are appropriate for which materials. Here we discuss them in more detail, moving from simple to complex. 5.5.1 Ideal Gases For ideal gases, we have Pv = RT . Ideal gases can be either calorically perfect or calorically imperfect . For all ideal gases, be they calorically perfect or calorically imperfect, it can be proved that the caloric equation of state takes on a simpler form: u = u ( T ) . (5.71) CC BY-NC-ND. 2011, J. M. Powers.

Subscribe to view the full document.

116 CHAPTER 5. THE FIRST LAW OF THERMODYNAMICS This will be proved in a later chapter. Now we can specify h for an ideal gas. From Eq. (5.52), h = u + Pv , and the ideal gas law, Pv = RT , we get h = u ( T ) + RT. (5.72) Thus, the enthalpy of an ideal gas is a function of T only: h = h ( T ) . (5.73) Now for the specific heats of an ideal gas, Eq. (5.63) gives c v ( T,v ) = ∂u ∂T vextendsingle vextendsingle vextendsingle vextendsingle v = d dT ( u ( T )) = c v ( T ) . (5.74) Separating variables in Eq. (5.74), we can also say for an ideal gas du = c v ( T ) dT. (5.75) For c P , Eq. (5.66) gives c P ( T,P ) = ∂h ∂T vextendsingle vextendsingle vextendsingle vextendsingle P = d dT ( h ( T )) = c P ( T ) . (5.76) Separating variables in Eq. (5.76), we get then dh = c P ( T ) dT. (5.77) Now, we can differentiate Eq. (5.72) to get dh = du + RdT. (5.78) Now substitute Eqs. (5.75,5.77) into Eq. (5.78) to get c P ( T ) dT = c v ( T ) dT + RdT, (5.79) c P ( T ) = c v ( T ) + R. (5.80) c P ( T ) c v ( T ) = R. (5.81) This is sometimes known as Mayer’s relation. Last, let us define the ratio of specific heats, k , as k = c P c v . (5.82) For general materials k = k ( T,v ). For an ideal gas, we have k = c v ( T ) + R c v ( T ) = 1 + R c v ( T ) . (5.83) So k = k ( T ) for an ideal gas. We will see that k ( T ) is often nearly constant. Since R> 0 and c v ( T ) > 0, we must have k> 1 for an ideal gas. In a later chapter, it will later be seen this result extends to general gases. CC BY-NC-ND. 2011, J. M. Powers.
5.5. CALORIC EQUATIONS OF STATE. 117 5.5.1.1 Calorically perfect A calorically perfect ideal gas (CPIG) has constant specific heat. Examples of CPIGs include noble and monatomic gases (e.g. He , Ne , Ar , O , H , N ) over a wide range of temperatures and pressures, and more complex molecules (e.g. O 2 , N 2 , CO 2 , CH 4 ) over narrower bands of temperatures and pressures. For the CPIG, c v is a constant, so ∂u ∂T vextendsingle vextendsingle vextendsingle vextendsingle v = c v . (5.84) But for the ideal gas, u = u ( T ), so the partial derivatives become total derivatives and du dT = c v . (5.85) Integrating, we get the simple caloric equation of state: u ( T ) = u o + c v ( T T o ) , (5.86) valid for CPIG. Note that u = u o + c v T c v T o , (5.87) u + Pv bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = h = u o + c v T c v T o + Pv bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright = RT , (5.88) h = u o + c v T c v T o + RT, (5.89) h = u o + RT o bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = h o + ( c v + R ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = c P T ( c v T o + RT o ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = c P T o , (5.90) h = h o + c P T c P T o . (5.91) So h ( T ) = h o + c P ( T T o ) , (5.92) valid for CPIG.

Subscribe to view the full document.

You've reached the end of this preview.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern