Question

yolume is vo, evaluate the constant. 25 A substance has an isothermal compressibility K aT3/p2 and an expansivity - bT2P where a and b are constants. Find the equation of state of the substance and ratio, ajb. 6From the equation of state given in Eq. (2-12) compute (a) the rate at which the gth of a rod changes with temperature when the tension is kept constant; (b) the rate

Answer 1

Isothermal compressibility, k = aT^3/p^2 expansivity, beta = bT2/p, a, b are constant According to formula, compressibility k = -1/v (Partialdifferential v/Partialdifferential p)_T = const Therefore aT^3/p^2 = -1/v (Partialdifferential v/Partialdifferential p)_cosnt T (Because It is isothermal) -1/v dv = aT^3/p^2 dp Integrating both side we get, -ln v = aT^3 0 epsilon x = e^aT^3/p (1) Again, expansivity beta = bT^3/p According to formula beta = 1/v (Partialdifferential v/Partialdifferential T)_p = const Therefore bT^2/p = 1/v (Partialdifferential v/Partialdifferential T)_const p (1/p bT^2) dT = 1/v dv On integrating both sides we get, 1/p bT^3/3 = ln v Equation of state v = e^bT^3/3p (2) \r\n Now, the equations of state obtained from compressibility and should be same at they are corresponding to a single substance. Therefore From equation (1) and (2) e^aT^3/p = e^bT^3/3p On comparing, a = b/3 Therefore Ratio a/b = 1/3
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