- Nomenclature
- Polytropic process
- Equation of State
- Relationships between P,V,T
- Adiabatic process
- Isothermal Process
- Internal Energy
- Constant Volume Process
- Constant Pressure Process

This page provides a limited notes on thermodynamic relationships useful to mechanical engineers.

Identifier | Description | Units (typical) |

c _{p} |
Specific Heat Capacity at Constant pressure | kJ/(kg K) |

c _{v} |
Specific Heat Capacity at Constant Volume | kJ/(kg K) |

P | Absolute Pressure | N / m ^{2} |

T | Absolute Temperature | K |

V | volume | m ^{3} |

m | mass | kg |

W | Work Output per unit mass | kJ/kg |

M | Molecular Weight | - |

R _{o} |
Universal Gas Constant = 8,31 | kJ /(kg mole.K) |

Q | Heat Quantity | kJ |

R | Gas Constant = R _{o} / M |
kJ /kg.K |

U | Internal energy (thermal) | kJ |

γ | Ratio c_{p} / c_{v} |
- |

The majority of frictionless processes for ideal gases are called polytropic processes and are in accordance with the following relationship

** PV ^{n} = constant
That is PV ^{n} = c therefore P = cV ^{-n}**

** PV = mRT**

Consider a piston in a frictionless cylinder

The work done on/by the gas in moving the piston δx = (PA)δx = P δ V = δ W

The gas is assumed to be expanding in balanced resisted reversible process.

The equation of state for an ideal gas is assumed to apply i.e PV = mRT

The total work done in moving the piston from state 1 to state 2 =

For a perfect gas - The relationship between Temperature , Pressure and Volume over a cycle

For an adiabatic process with no transfer of heat across the system boundary.(Q = 0 )

Consider a fixed mass of gas in a cylinder which is expanding in a reversible manner...

For an adiabatic process there is no heat transfer.

Therefore applying the
first law of thermodynamics ..heat transfer (= 0) = increase in internal energy +external work
done by gas...

Therefore the increase in internal energy = - External work done by gas

It is shown below that c_{p} - c_{v} = R = c_{v} ( c_{p} / c_{v} -1) and therefore

γ = 1.4 for Air, H _{2},
O _{2}, CO, NO, Hcl

γ = 1.3 for CO _{2}, SO _{2},
H _{2}O, H _{2}S, N _{2}O, NH _{3}, CL _{2},
CH _{4}, C _{2}H _{2}, C _{2}H _{4}

In a isothermal process the temperature = constant and therefore

** PV = c and P = c / V **

Although it is not possible to determine the absolute value of the internal energy
of a substance. The internal energy change between the initial and final
equilibrium states of any process is definite and determinable.

It can be easily proved that the internal energy of a fluid depends on the temperature
alone and not upon changes in the pressure or volume.

If a definite mass of gas (m) at constant volume is a closed system is heated from
initial conditions P_{1}, V, T_{1}, U_{1} to
P_{2}, V , T_{2},U_{2}. As the volume is fixed then no work
has been done. Then in accordance with the First Law
of Thermodynamics (δQ = δU + δW ).

**mC _{v} (T_{2} - T_{1}) = (U_{2} - U_{1}) + 0 **

or U_{2} - U_{2} = mC_{v} (T_{2} - T_{1})

If a definite mass of gas (m) at constant volume is a closed system is heated from
initial conditions P, V_{1}, T_{1}, U_{1} to
P, V_{1} , T_{2},U_{2}. As the volume is fixed then no work
has been done. Then in accordance with the First Law
of Thermodynamics (δQ = δU + δW ).

**mc _{p} (T_{2} - T_{1}) **

= (U_{2} - U_{1}) + P (V_{2} - V_{1})

= (U_{2} - U_{1}) + mR (T_{2} - T_{1})

mc_{v} (T_{2} - T_{1}) = U_{2} - U_{1} therefore
mc_{p} (T_{2} - T_{1}) = mc_{v} (T_{2} - T_{1}) + mR (T_{2} - T_{1}) therefore

** c _{p} = c_{v} + R... and ..**

c_{p} - c_{v} = R = PV/mT

- Thermodynamics..NASA - Glenn Research center at Series of informative notes on Thermodynamics
- Thermodynaic properties,property relationships and processes..A very detailed clear study of the subject, (3,3 Mbyte download)