Introduction
This page provides a limited notes on thermodynamic relationships useful
to mechanical engineers.
Nomenclature
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} 
 
Thermodynamic Process Relationships
General Polytropic Process
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}
Equation of state for and Ideal Gas
PV = mRT
Thermodynamic Relationships between P,V & T
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
Adiabatic Process.
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}
Isothermal Process
In a isothermal process the temperature = constant and therefore
PV = c and P = c / V
Internal Energy, C_{p} and 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.
Heating at constant 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})
Heating at constant pressure....
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
