We emphasize that in order to obtain work and energy

This preview shows page 4 - 5 out of 5 pages.

where we have used the ideal gas law in the last step. We emphasize that, in order to obtain work and energy in Joules, pressure should be in Pascals (N / m 2 ) and volume should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5. (a) The internal energy change is
Image of page 4

Subscribe to view the full document.

  3 3 3 3 int int 3 5 5 2.0 10 Pa 4.0m 5.0 10 Pa 2.0m 2 2 5.0 10 J. c a c c a a E E p V p V   (b) The work done during the process represented by the diagonal path is  3 3 diag = 3.5 10 Pa 2.0m 2 a c c a p p W V V which yields W diag = 7.0×10 3 J. Consequently, the first law of thermodynamics gives 3 3 3 diag int diag ( 5.0 10 7.0 10 ) J 2.0 10 J. Q E W     (c) The fact that E int only depends on the initial and final states, and not on the details of the ―path‖ between them, means we can write 3 int int int 5.0 10 J c a E E E   for the indirect path, too. In this case, the work done consists of that done during the constant pressure part (the horizontal line in the graph) plus that done during the constant volume part (the vertical line):  3 3 4 indirect 5.0 10 Pa 2.0m 0 1.0 10 J. W Now, the first law of thermodynamics leads to 3 4 3 indirect int indirect ( 5.0 10 1.0 10 ) J 5.0 10 J. Q E W     11. Suppose the gas expands from volume V i to volume V f during the isothermal portion of the process. The work it does is i f V V V V V V nRT V dV nRT pdV W f i f i ln , where the ideal gas law pV = nRT was used to replace p with nRT / V . Now V i = nRT / p i and V f = nRT / p f , so V f / V i = p i / p f . Also replace nRT with p i V i to obtain ln . i i i f p W pV p Since the initial gauge pressure is 1.03 10 5 Pa, p i = 1.03 10 5 Pa + 1.013 10 5 Pa = 2.04 10 5 Pa. The final pressure is atmospheric pressure: p f = 1.013 10 5 Pa. Thus  5 5 3 4 5 2.04 10 Pa 2.04 10 Pa 0.14m ln 2.00 10 J. 1.013 10 Pa W During the constant pressure portion of the process the work done by the gas is W = p f ( V i V f ). The gas starts in a state with pressure p f , so this is the pressure throughout this portion of the process. We also note that the volume decreases from V f to V i . Now V f = p i V i / p f , so . 10 44 . 1 ) 14 . 0 )( 10 04 . 2 10 013 . 1 ( ) ( 4 3 5 5 J m Pa Pa V p p p V p V p W i i f f i i
Image of page 5

{[ 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