By Bernard Morrill
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Extra resources for An Introduction to Equilibrium Thermodynamics
0) — - Ζ\Δ) Κ t2 = 1660°F Since the process is a constant pressure then CP is heat capacity desired. C » = JT 1-10 = 212^530 = 0 0 6 29 B T U / RO ENERGY OF A PERFECT GAS (Translational Degrees of Freedom) It will be very convenient, at this point, to be able to calculate the internal energy of a perfect gas. Knowing the energy of the gas will enable us to predict the t h e r m o d y n a m i c p r o p e r t i e s t . It is not possible to uniquely derive from a purely macroscopic point of view the energy of a perfect gas.
In effect these foregoing assumptions define a perfect m o n a t o m i c gas. 1) are not derivable from this equation since it is not a fundamental equation. Energy of a Perfect Gas 27 ζ m vi y χ Fig. 11 the internal energy since there is no contribution of kinetic and potential energies at the c e n t e r of mass of the system. A n additional assumption is necessary and that is that the gas is unbiased; meaning that the velocities and therefore the kinetic energies in each of the three orthogonal directions, JC, y, and z, are statistically the same.
A hsat source is brought to the piston and remains in contact until the 3 piston encloses a volume of 5 ft . Calculate the work done by the system during the expansion process. 3 1-8. The piston of a cylinder moves so that the volume expands from 3 ft to 3 5 ft while the temperature of the piston-cylinder system remains constant. If the pressure is 200 psia initially, how much work is done during the expansion process? 1-9. Determine whether the following expressions possess exact derivatives: 2 (a) ζ = 3xy + 4x 2 (b) ζ = x + 3xy + yjt 3 (c) ζ = î ( y cosxdx +sinxdy) 1-10.
An Introduction to Equilibrium Thermodynamics by Bernard Morrill