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Materials/Solid State Physics Problem Check.

  1. May 17, 2006 #1
    Hello, I've been looking through some books and done some examples. Unfortunately, they never give proper solutions. If anyone could help verify my responses, then it would be much appreciated.

    1.The Van der Waals constants for a gas are found to be a=10m^3.Pa/mol, and b=0.001m^3/mol. Calculate the critical temperature in K to the nearest degree.

    T(c)=8*10 / 27*0.001*8.314 = 356K

    2. At a pressure of p=2atm, the melting point of a material is T(m)=370K. The density of the solid phase is rho(s)=5400kg/m^3 and the density of the liquid phase is rho(l)=4500kg/m^3. If the pressure is changed to p=57atm, the fractional change in transition temperature is (deltaT)/T=0.045. Calculate the latent heat of fusion in kJ/kg to one decimal place.

    (deltaP)/(deltaT) = (deltaH) / T(deltaV)
    => (deltaH) = (deltaP)T(deltaV)/(deltaT)
    deltaP = 57-2 = 55atm = 55*101325Pa = 5572875Pa
    deltaV = rho(l)^-1 - rho(s)^-1 = 1/27000 m^3/kg
    deltaT = 0.045T = 0.045*370 = 16.65
    .'. (deltaH) = 5572875*370/(16.65*27000) = 4586.7J/kg = 4.6kJ/kg

    3. For a transition which occurs at a temperature of T(m) 150degrees C with a latent heat of 500kJ/mol, calculate the change in entropy in kJ/(mol.K) to one decimal place.

    deltaS = deltaH/T = 500000/423 = 1.2kJ/(mol.K)

    4. It is observed that at a particular temperature and pressure a mixture has 2 solid phases and 2 liquid phases in equilibrium. If the system is invariant at this temperature and pressure, calculate the number of components in the mixture.

    F = C+2
    Maximum degrees of freedom is 2+2 = 4
    C = components
    => 4-2 = C
    .'. C = 2

    5. A bar of material is stretched from it's initial length of 3m to 3.001m. If the bar has a 5cm*5cm cross section, and it's Young's modulus is 7.6GPa, then assuming elastic behaviour, calculate the applied force to the nearest N.

    Strain = Change in Length/ Initial Length = 0.001/3
    Stress = Young's Modulus*Strain = 7.6e9 * 0.001/3 = 2533333.333.......
    Force = Area*Stress = 0.05m*0.05m * 2533333.333.....
    .'. F = 6333N

    6. A cylindrical rod of diameter 1cm is stretched from it's initial length of 1.2m to a length of 1.23m. If the diameter of the bar is reduced to 0.9966cm, then assuming elastic behaviour, calculate Poisson's ratio for the material to 2 decimal places.

    mu = poisson's ratio = - (strain in x)/(strain in z)
    mu = -(change in diameter/diameter)/(change in length/length)
    mu = -(-0.0034e-2/1e-2)/(0.03/1.2) = +0.136 = 0.14

    7. A material has a bulk modulus of 56GPa and a Poisson's ratio of 0.27. Calculate Young's Modulus to the nearest GPa.

    Bulk Modulus = K = Young's Modulus/(3-6mu)
    Y = K(3-6mu) = 56e9(3-6(0.27)) = 77.3GPa = 77GPa

    I have some more problems which I've done but I think that may be enough for now. I hope I have gotten these all right but sometimes I make stupid mistakes. Thanks in advance.

  2. jcsd
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