Thermodynamic properties, Study notes of Chemistry

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THERMODYNAMIC

PROPERTIES AND

CALCULATION

Academic Resource Center

THERMODYNAMIC PROPERTIES

A quantity which is either an attribute of an entire system or is a function of position which is continuous and does not vary rapidly over microscopic distances, except possibly for abrupt changes at boundaries between phases of the system; examples are temperature, pressure, volume, concentration, surface tension, and viscosity. Also known as macroscopic property.

BASIC CONCEPTS-

 PV diagram

 Virial Equations of State

PV = a + bP + cP^2 + ...... →

 Ideal gas: Z=1 or PV = RT

 Van Der Waals Equation of State

For Ideal Gas: Equation for Calculation

Heat capacity:

dQ + dW = CvdT dW = – PdV dQ = CvdT + PdV

Let V=RT/P :

PRIMARY THERMODYNAMIC PROPERTIES—

P, V, T, S & U

Combining the first and second laws in reversible process The only requirements are that the system be closed and that the change occur between equilibrium states.  H = U + PV  A = U – TS  G = H – TS d(n U ) = Td(nS) – Pd(nV) d(nH) = Td(nS) + (nV)dP d(nA) = – Pd(nV) – (nS)d d(nG) = (nV)dP – (nS)dT

dU = TdS – PdV dH = TdS + VdP dA = – PdV – SdT dG = VdP – SdT

Maxwell’s equation

KEYS

SOLUTIONS:

 In suche case take the system as 1 mol of air contained in an imaginary piston/cyclinder arrangement. Since the processes considered are mechanically reversible, the piston is imagined to move in the cylinder withour friction. The final volume is

 (a) During the first step the air is cooled at the constant pressure of 1 bar until the final volume of 0.004958 m^3 is reached. The temperature of the air at the end of this cooling step is:

SOLUTIONS

 also applies to the entire process. But

 and therefore,

 Hence ,

 and

SOLUTIONS

 Two different steps are used in this case to reach the same final state of the air. In the first step the air is heated at a constant volume equal to its initial valve until the final pressure of 5 bar is reached. The air temperature at the end of this step is:

 For this step the volume is constant, and

 During the second step the air is cooled at the constant pressure of 5 bar to its final state:

EXAMPLE 2

 Air is compressed from an initial condition of 1 bar and 25 ℃ to a final state of 5 bar and 25℃ by three different mechanically reversible processes in a closed system:

 (a) Heating at constant volume followed by cooling at constant pressure.

 (b) Isothermal compression.

 (c)Adiabatic compression followed by cooling at constant volume.

 Assume air to be an ideal gas with the constant heat capacities, CV= (5/2)R and CP = (7/2)R. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process.

KEYS

SOLUTIONS

 (a) The heat transferred, from Ex.1(b) is Q=- 9.915J. Thus by the first law applied to the entire process:

 (b) Equation for the isothermal compression of an ideal gas applies here:

 (c) The intial adiabatic compression of the air takes it to its final volume of 0.004958m^3. The temperature and pressure at this point are:

SOLUTIONS

T 2 = T 1

V 1

V 2

γ− 1

)^0.^4 = 567. 57 K

P 2 = P 1

V 1

V 2

γ

)^1.^4 = 9. 52 bar

For this step Q=0, and

W = CV ∆T = 20. 785 567. 57 − 298. 15 = 5 , 600 J

For the second step at constant V, W=0. For the overall process,

W = 5 , 600 + 0 = 5 , 600 J