{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

general chemistry4 - 5.60 Spring 2004 Lecture#6 page 1...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
5.60 Spring 2004 Lecture #6 page 1 Reversible Adiabatic Expansion (or compression) of an Ideal Gas one mole ideal gas reversible adiabatic 1 mol gas ( V 1 , T 1 ) = 1 mol gas ( V 2 , T 2 ) Adiabatic dq = 0 Reversible dw = -pdV Ideal gas dU = C v dT \ From 1 st Law dU = -pdV C v dT = -pdV along path C V dT = - pdV p = RT V C V dT T = - R dV V Integrating: C V dT T T 1 T 2 = - R dV V V 1 V 2 C V ln T 2 T 1 = - R ln V 2 V 1 T 2 T 1 C V = V 1 V 2 R T 2 T 1 = V 1 V 2 R C V C p - C V = R for i.g. æ ææææ æ T 2 T 1 = V 1 V 2 C p C V - 1 Define g C p C V T 2 T 1 = V 1 V 2 g - 1 For monatomic ideal gas: C V = 3 2 R C p = 5 2 R g = 5 3 ( > 1 generally) Since g > 1 g - 1 > 0
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
5.60 Spring 2004 Lecture #6 page 2 In an adiabatic expansion ( V 2 > V 1 ), the gas cools ( T 2 < T 1 ). And in an adiabatic compression ( V 2 < V 1 ), the gas heats up. For ideal gas (one mole) T = pV R p 2 p 1 = V 1 V 2 g In other words p 1 V 1 g = p 2 V 2 g pV g is constant along a reversible adiabat For isothermal process T = constant pV = constant pV is constant along an isotherm V 2 adiabat < V 2 isotherm because the gas
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern