Gas Laws Laws 1

Technology, Engineering and Environment Engineering Design and Manufacturing Systems

Applied Thermodynamics Ideal Gas Laws

Combined Gas Law For a fixed mass of ideal gas undertaking a process from state 1 to state 2:

2

22

1

11

T

Vp

T

Vpm …………………………….Eq 1

Characteristic Equation of State For unit mass of an ideal gas:

RT

pv

where: p = pressure [N/m2] v = volume for a mass of 1 kg [m3] T = absolute temperature [K] R = characteristic gas constant [J/kg K], dependent on the type of gas. And so for mass m of ideal gas:

mRT pV …………………………….Eq 2

where: V = volume of mass m kg of gas [m3] m = mass [kg]

Universal Gas Constant A kmol of any ideal gas at a given temperature and pressure will have the same volume, ie:-

where: = universal gas constant [8.314 kJ/kmolK] = molar volume [m3/kmol] so for any volume of ideal gas:-

where n = amount of gas [kmol] V = volume [m3] from equations 2 and 3, and substituting m = n M, M = molar mass [kg/kmol] leads to:-

where: R = characteristic gas constant [J/kg K]

Gas Laws Laws 2

Avogadro's Law Equal volumes of different gases at the same temperature and pressure have the same number of particles (i.e. molecules). Kilogram-mol The kilogram-mol, usually shortened to kgmol or kmol, is that amount of substance which contains as many particles as there are atoms in 12 kg of the isotope Carbon-12. The number of particles per kmol is 6.022 x 1026, known as Avogadro's constant. Relative molecular mass In the following notes the word molecule is used to include those substances where the particle happens to be an atom. Different gases have molecules of different relative molecular mass. The relative molecular mass Mr is defined as the ratio of the mass of the molecule divided by 1/12 of the mass of the carbon atom. Example relative molecular masses:- monotomic:-Mr triatomic:- Mr Argon (Ar) 40 Carbon dioxide (CO2) 44 Helium (He) 4 Sulphur dioxide (SO2) 64 diatomic:- hydrocarbons:- Carbon monoxide (CO) 28 Ethane (C2H6) 30 Hydrogen (H2) 2 Methane (C H4) 16 Nitrogen (N2) 28 Propane (C3H8) 44 Oxygen (O2) 32 A kmol of a gas is that quantity equal to the relative molecular mass expressed in kg. Hence a kmol of Nitrogen is 28 kg of Nitrogen; a kmol of Methane is 16 kg of Methane.

Gas Laws Laws 3

Polytropic process The pressure : volume relationship for many typical expansion (Fig. 3a) and compression (Fig. 3b) processes for gases can be defined by a polytropic process:

constantpVn

that is, for a polytropic process from state 1 to state 2:

n

22

n

11 VpVp ..…..……………………….Eq 5

where: n = polytropic index. The value of n depends on the type of process and there are a number of special cases as described in the following notes. Combining Eq 1 and Eq 5 leads to the following equation applicable to polytropic processes:-

1

2

1

1

1

2

1

2

V

V

p

p

T

Tn

n

n

…………………………….Eq 6

Special cases of the polytropic process isentropic process An isentropic (i.e. constant entropy) process is one that is frictionless and adiabatic. It is a special case of the polytropic process for which the index n = :-

pV = c where c = constant = the ratio of specific heats cp/cv and for air = 1.4 The property entropy (symbol S) remains constant in isentropic expansion (Fig. 3c) and compression (Fig. 3d) processes. Refer to the separate note for a brief introduction to the property entropy. Special cases of the polytropic process isothermal process

An isothermal (constant temperature) process is a special case of the polytropic process for which the index n = 1, that is :-

pV = c where the constant c = mRT

Gas Laws Laws 4

During an isothermal expansion process, Fig. 3a, heat is supplied to the system. The entropy of the gas increases, Fig 3c, by an amount given by the ratio of heat transferred divided by the absolute temperature. For an isothermal compression, heat has to be transferred from the gas and so the entropy of the gas is reduced, Fig 3d. Polytropic, isentropic and isothermal expansion processes on p-v and T-s diagrams: p 1 T 1 isothermal 2 isothermal polytropic isentropic 2 isentropic 2 2 polytropic 2 2 v s Fig. 3a Fig. 3c Polytropic, isentropic and isothermal compression processes on p-v and T-s diagrams: 2 p 2 isentropic T 2 isentropic 2 polytropic 2

polytropic

isothermal 2 1

1 isothermal

v s

Fig. 3b Fig. 3d It can be seen that the isentropic and isothermal processes represent two extremes of the general polytropic process. An isentropic process, that is a frictionless compression or expansion process, during which there is no heat transfer, can be approached by real compression/expansion processes if carried out very quickly. Thus, for a simple theoretical model of a high speed reciprocating engine cycle, the compression and expansion processes can be considered isentropic. An isothermal process can be approached for real compression and expansion process only if carried out very slowly, thus allowing time for heat transfer to occur

Gas Laws Laws 5

and the temperature to remain constant at the value of the surroundings. High speed reciprocating engine cycle compression and expansion processes are not isothermal because they are too quick and therefore do not allow time for external heat transfer to occur. Real compression and expansion processes typically fall between the two extremes of isothermal and isentropic. Such processes are called polytropic and have an index n which falls between the values of 1 and . For air, the value of is1.4 . Work transfer for polytropic process The magnitude of the work transfer for a frictionless compression or expansion process is given by the area under the p-v diagram:

2

1

dV pW

For the polytropic process, p = c/vn. Substituting and integrating between state 1 and 2 gives:

n-1

)T-mR(T

1-n

Vp-VpW 122211 …………………………….Eq 7

If W is positive, it indicates that work is being done by the system, i.e. an expansion process . If W is negative then work is being done on the system, i.e. a compression process Work transfer for isentropic process The work transfer for an isentropic process is given by the same expressions as for a polytropic process but with substituted in place of the polytropic index n. Work transfer for isothermal process As before, the magnitude of the work transfer is given by the area under the p-V diagram:

2

1

dV pW

Gas Laws Laws 6

For the isothermal process, substituting p = c/v and integrating between states 1 and 2 leads to the an expression for the isothermal special case:

1

2e11

V

Vlog VpW …………………………….Eq 8

Since for an isothermal process: pV = mRT and: p1V1 = p2V2 the work transfer can be expressed as:

2

1e

1

2e

p

plog mRTor W

V

Vlog mRTW

Constant volume and constant pressure processes Constant volume and constant pressure processes are also special cases of the polytropic process, pvn = c, with n = 0 and n = respectively. Figs. 4a and 4b show constant volume and constant pressure processes for heat addition from state 1 to state 2: p T 2 2 2 constant constant volume volume constant pressure 1 constant 2 1 pressure v s Fig. 4a Fig. 4b Constant volume processes Consider the system diagram and p-v and T-s diagrams Figs. 4a and 4b: Q

Gas Laws Laws 7

The heat transfer to raise the temperature of a quantity of gas whilst maintaining constant volume is given by:

ΔTmcQ v …………………………….Eq 9

Consideration of the non-flow energy equation: Q = W + U, in which W is zero for a constant volume process, shows that the heat transfer for a constant volume process is also the change of internal energy, and so:

Tmc ΔU v …………………………….Eq 10

Note that the change of internal energy for an ideal gas is always given by Eq 10, whatever the process. Constant pressure processes Consider the system diagram and p-v and T-s diagrams Figs. 4a and 4b: Q

W The heat transfer to raise the temperature of a quantity of gas whilst maintaining constant pressure is given by:

ΔTmcQ p …………………………….Eq 11

Applying the non-flow energy equation:

ΔUWQ

Since TmR)V-p(V Wprocess pressureconstant afor and TmcU 12v , then:

TmR TmcTmc vp

Gas Laws Laws 8

from which:

Rcc vp …………………………….Eq 12

Work and heat transfer for steady flow processes with ideal gases Work and heat transfer for a steady flow processes are related by the steady flow energy equation to the change in enthalpy of the fluid (refer to separate hand-out). Neglecting kinetic and potential energy terms, the SFEE reduces to: :

h1 + Q = h2 + W where h, the property specific enthalpy, is defined by:

h = u + pv Now for an ideal gas:

p v = RT and u = cv T so for unit mass of an ideal gas:

h = cv T + RT

and since:

R = cp - cv

then:

h = cp T or Δh = cp ΔT Calculations for air

Air can be considered as an ideal gas for approximate calculations and the following values can be used with reasonable accuracy over a wide range of temperature and pressure:

cv = 0.718 kJ/kgK cp = 1.005 kJ/kgK R = 0.287 kJ/kgK = 1.4