By Michael J.R. Hoch
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Extra resources for Statistical and Thermal Physics : An Introduction
Example text
V With the assumption that ∆E ≈ 0, we have ∆Q = ∫ Vfi P dV. For a van der Waals gas, with n = 1 mol, P = RT/(V − b) − a/V 2. This gives ∫ Vf Vi P dV = RT ∫ Vf Vi dV −a V −b ∫ Vf Vi dV ⎛ V − b⎞ ⎛ 1 1⎞ = RT ln ⎜ f + a ⎜ − ⎟ . 9) ⎝ Vi − b ⎟⎠ ⎝ Vf Vi ⎠ V2 In this case, it is necessary to know the van der Waals constants a and b to obtain ∆Q. For Vf < Vi, the first term will be negative and the second positive. This reflects the fact that in the pressure term, which involves the constant a, a decrease in volume increases the importance of the interparticle attractive interaction, whereas in the volume term, the constant b simply reduces the effective volume available to the particles.
6 represents a collision process in a container. As an idealization, assume that all collisions are elastic; the change in the momentum component px for a particle i in striking the smooth wall, as shown, is Δpix = 2mvix. 6, the number of molecules n that strike area A in a time t is given by the ratio of the sum, for all particles i, of the volumes of cylinders of length vix t and cross-sectional area A to the total volume or n = 12 ( A/V )∑ iN=1 vix t . The factor 12 is introduced because on average half the molecules in each small volume considered will have a velocity component in the + x direction whereas the other half will have a component in the - x direction.
The ideal gas equation of state may be applied at all stages of the process. 3 depicts the situation. 7) is given by dE = dQ − P dV. 12. Because T remains fixed, the average energy of the molecules does not change. With dE = 0, we have dQ = P dV, f and integration leads to ∆Q = ∫ i P dV. The ideal gas equation gives P = nRT/V, and we obtain ΔQ = nRT ∫ Vf Vi dV ⎛V ⎞ = nRT ln ⎜ f ⎟ . 8) If Vf < Vi , it follows that ∆Q < 0, and this shows that heat is rejected by the system in the compression process.
