![]() ![]() Using W = − HdM, we obtain W =−Ī log log plot of mass vs. R ( + for lower branch, − for upper branch.). Solve for the magnetic field: H = ±H0 + tanh−1 (M/M0 ) Thus the work on the system is Zġ.7 Consider the hysteresis cycle in the sense indicated in Fig.1.6. ![]() Therefore T =ġ.6 R Work done by the system is − HdM. That is, heat just pass from one body to the other. Therefore total heat absorbed ∆Q = ∆Q1 + ∆Q2 = 0. ![]() At constant T, we have dp/P = dn/n.Therefore n(z) = n(0)e−mgz/kB Tġ.5 No change in internal energy, and no work is done. For equilibrium, the weight must equal the pressure differential: dP = −gdM. The weight of the element is −gdM, where dM is the mass of the element: dM = mndz, where m is the molecular mass, and n = P/kB T is the local density, with P the pressure. ![]() N kB T N kB T0b 1−b T = V V0 Work done = P ∆V = bN kB ∆T Pġ.4 Consider an element of the column of gas, of unit cross section, and height between z and z+dz. Heat absorbed equals total work done, since internal energy is unchanged in a closed cycle. The total work done is the sum of the above. 1.2 Work done along various paths are as follows ab: Z b Z b Vb dV P dV = N kB T1 = N kB T1 ln V V a a a cd: Introduction to Statistical Physics Solution Manual Kerson Huangġ.1 Mass of water =106 g, temperature raised by 20◦ C. ![]()
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