1. Application of Environment Spatial Information System HW – Perfect Gas Minkasheva Alena Thermal Fluid Engineering Lab. Department of Mechanical Engineering Kangwon National University HOMEWORK

2. Contents Definition of Perfect GasDefinition of Perfect Gas The Gas LawsThe Gas Laws Perfect Gas EquationPerfect Gas Equation Perfect Gas Equation of StatePerfect Gas Equation of State Perfect GasesPerfect Gases Perfect Gas WorkPerfect Gas Work Summary of WorkSummary of Work Molar Specific Heat of Perfect GasMolar Specific Heat of Perfect Gas

3. Definition of Perfect Gas Perfect gas:Perfect gas: “gas where intermolecular forces are negligible”. It acts as a continuous material in which the properties are determined by statistical average of the particle effects. At low gas densities, all gases can be treated as perfect gases. A gas that is not perfect is called a real gas. Avogadro’s number N A = 6.02·10 23 mol -1 Number of moles in a sample n = N/N A Universal Gas Constant R = 8.31 J/mol·K Boltzmann constant k = R/N A = 1.38 x 10 −23 J/K

4. The Gas Laws Boyle’s Law: For a fixed amount of gas and constant temperature, PV = constant Charles’s Law: At constant pressure the volume is linearly proportional to temperature V/T = constant Avagadro’s Law: For a fixed pressure and temperature, the volume of a gas is directly proportional to the number of moles of that gas V/n = k = constant

5. Perfect Gas Equation Perfect Gas Law: Perfect Gas Law: the functional relationship between the pressure, volume, temperature and moles of a gas. PV =nRT Each of the individual laws is contained in this equation: Boyle's Law: Boyle's Law: PV = k 1 = nRT Charles's Law: Avagadro’s law: When any of the other three quantities in the ideal gas law have been determined the last one can be calculated.

6. Perfect Gas Equation of State Thermodynamic state equation: P = absolute (not gauge) pressure in N/m 2 (or Pascals) V = volume in m 3 N = number of molecules T = absolute temperature in K k = Boltzmann’s constant = 1.38·10 -23 J/K n = number of moles R = universal gas constant = 8.31 J/mol/K If n is constant, then: ρ = density in kg/m 3

7. Perfect Gases Consider special cases: Pressure: Pressure: Isobaric - constant pressure Volume: Volume: Isochoric (or isovolumic) - constant volume Temperature: Temperature: Isothermal - constant temperature For normal air:

8. Perfect Gas Work Three variables in the ideal gas law (with n being constant). Consider special cases: Pressure: Pressure: Isobaric - constant pressure i f Volume Pressure 1

9. i f Volume Pressure Three variables in the ideal gas law (with n being constant). Consider special cases: Volume: Volume: Isochoric (or isovolumic) – constant volume since the integral limits are equal Perfect Gas Work 2

10. i f Volume Pressure Three variables in the ideal gas law (with n being constant). Consider special cases: Temperature: Temperature: Isothermal - constant temperature Gas expands from V i to V f, P = nRT/V Perfect Gas Work 3

11. Work done at constant pressure P is constant, W = P(V f – V i ) = PΔV Work done at constant volume dV = 0, so W = 0 Work done by ideal gas at constant temperature Summary of Work

12. Molar Specific Heat of Perfect Gas Molar specific heat: Q = cn (T f – T i ) The specific heat c is a value that depends on the ability of a substance to absorb energy. As such, c depends on both the type of material and whether the process is a constant volume process or a constant pressure process. 1) Constant-volume process 2) Constant-pressure process 3) Arbitrary process i f Volume Pressure f f 3 2 1 T+ΔT T 1

13. Molar Specific Heat of Perfect Gas Molar specific heat at constant volume: C V Q = n CV ΔT (constant V process) since V = const, so W = 0, so ΔE int = Q – W = n CVΔT or E int = n CV T For ideal gas, the change in internal energy depends only on the change in gas temperature. i f Volume Pressure f f 3 2 1 T+ΔT T 2

14. Molar Specific Heat of Perfect Gas Molar specific heat at constant volume: C P Q = nC P ΔT (constant P process) ΔE int = Q – W, since P = const, W = PΔV = nRΔT ΔE int = nCVΔT = nC P ΔT – nRΔT = n(C P – R)ΔT Therefore C V = C P – R

15. THE END