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ENGR5003 / Week 6
ENGR272 week 6 mock Q1Past paperCurrent spec2:150 marks60 min

ENGR272 Week 6 mock exam Q1 (Dr Andalibi)[VALID] official answers

A pure substance is modelled by the van der Waals equation of state:

P=RuTv−b−av2P = \dfrac{R_u T}{v - b} - \dfrac{a}{v^2}

where aa and bb are constants, vv is the molar volume, and RuR_u is the universal gas constant. You may use the Maxwell relations provided in the formula sheet.

Formulas you may need
  • Exact differential: dz=M dx+N dydz = M\,dx + N\,dy with (∂M∂y)x=(∂N∂x)y\left(\dfrac{\partial M}{\partial y}\right)_x = \left(\dfrac{\partial N}{\partial x}\right)_y (on the formula sheet)
  • Reciprocity: (∂x∂z)y=1(∂z/∂x)y\left(\dfrac{\partial x}{\partial z}\right)_y = \dfrac{1}{(\partial z/\partial x)_y}; cyclic relation (∂x∂y)z(∂y∂z)x(∂z∂x)y=−1\left(\dfrac{\partial x}{\partial y}\right)_z\left(\dfrac{\partial y}{\partial z}\right)_x\left(\dfrac{\partial z}{\partial x}\right)_y = -1 (on the formula sheet)
  • Gibbs relation dg=v dP−s dTdg = v\,dP - s\,dT (on the formula sheet; also given in the question)
  • Maxwell relation (∂s∂P)T=−(∂v∂T)P\left(\dfrac{\partial s}{\partial P}\right)_T = -\left(\dfrac{\partial v}{\partial T}\right)_P (on the formula sheet)
  • van der Waals equation P=RuTv−b−av2P = \dfrac{R_uT}{v - b} - \dfrac{a}{v^2} (given in the question; otherwise learn this)
  1. (a)
    Starting from the Gibbs relation dg=v dP−s dTdg = v\,dP - s\,dT, derive an expression for (∂s∂P)T\left(\dfrac{\partial s}{\partial P}\right)_T in the form of a Maxwell relation. Explain briefly why this derivative is useful in thermodynamic property calculations.
    [10]
  2. (b)
    Using the van der Waals equation of state, determine an explicit expression for (∂v∂T)P\left(\dfrac{\partial v}{\partial T}\right)_P in terms of specific volume and temperature as the state variables. Show all steps clearly.
    [12]
  3. (c)
    Using your results from parts (a) and (b), obtain an expression for (∂s∂P)T\left(\dfrac{\partial s}{\partial P}\right)_T for a van der Waals gas.
    [10]
  4. (d)
    Consider a van der Waals gas with a=0.90 Pa m6 mol−2a = 0.90\ \mathrm{Pa\,m^6\,mol^{-2}}, b=4.0×10−5 m3 mol−1b = 4.0 \times 10^{-5}\ \mathrm{m^3\,mol^{-1}}, Ru=8.314 J mol−1 K−1R_u = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}. At T=400 KT = 400\ \mathrm{K} and P=5.0 MPaP = 5.0\ \mathrm{MPa} the molar volume is v=4.86×10−5 m3 mol−1v = 4.86 \times 10^{-5}\ \mathrm{m^3\,mol^{-1}}. Calculate the numerical value of (∂s∂P)T\left(\dfrac{\partial s}{\partial P}\right)_T at this state. Give your answer in units J mol−1 K−1 Pa−1\mathrm{J\,mol^{-1}\,K^{-1}\,Pa^{-1}}.
    [10]
  5. (e)
    Comment briefly on the physical meaning of the sign of (∂s∂P)T\left(\dfrac{\partial s}{\partial P}\right)_T for real gases.
    [8]