# The Ideal Gas Equation

The relationships between volumes of gases under constant temperature and pressure were established by Boyle, Charles.

Boyles law was first published in 1662. It states

The pressure times the volume of a gas is equal to some constantk.PV=k.

Charles/Gay-Lussac Law, states that

At constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature (in kelvin) increases or decreases.

*P*_{1}/*T*_{1} = *P*_{2}/*T*_{2}

Both of these laws can be combined to give an overall equation which describes the behaviour of gases.

*P*_{1}*V*_{1}/*T*_{1} = *P*_{2}*V*_{2}/*T*_{2}

Avogadro's Law, states that the volume *V* is related to the number of moles of gas, *n* by

*V*/*n* = *k*

*P*_{1}*V*_{1}/*T*_{1}*n* = *P*_{2}*V*_{2}/*T*_{2}*n* = k, where k is a constant.

Which is more familiar as

*PV*= *nRT*

The constant *k* has been replaced by *R* the molar gas constant.

Where *P* is the pressure, *V* is the volume, *n* is number of moles of gas present, *R* is the molar gas constant has a value very close to 8.31 m^{2} kg s^{-2} K^{-1} mol^{-1} and *T* is the temperature in Kelvin.

## The PV diagram

We can plot the path of the pressure against the volume, if we consider the change in the state variables to be quasi-static.

For a constant temperature the pressure against volume curve looks like

### Isothermal Process

Example1: Boiling of water in the open air. In general most isobaric phase changes are isothermal. In this example the system does work as the steam-produced pushes against the atmosphere as it expands. Neither the heat Q , the work W, or the change in internal energy DU are zero. In this case Q = mLv since the water changes phase. Example 2: In general for an Ideal gas U is only a function of the temperature so that DU is always equal to zero for an isothermal process.Since DU = 0 then W = Q from the First Law. What has to happen for this process to be isothermal is that the gas in a cylinder is compressed slowly enough that heat flows out of the gas at the same rate at which is being done on the gas. The ideal gas law can be used to determine the work done W = PV ln(Vf/Vo) which is also the equation for Q. Note that P1V1 = P2V2 = nRT, the ideal gas law for an isothernal process.

### Adiabatic Process

Example Compression of a Gas in an Insulated Cylinder. In this case any change in the internal energy of the gas is due to work done on it or by it, DU = W. Normally if DU changes the temperature of a system will change. Any temperature rise or fall is due to the work done or by the gas alone and not due to heat flowing into or out of the system since Q = 0. If a process is carried out fast enough the heat flow will be small and the process can be approximate as being adiabatic. This happen because heat flow is in general a slow process. Observe that we did not say that Q is constant because it not a state variable. Q represent an energy transfer not the heat energy of the system. In addition to the ideal gas law PV = NkT, the quantity PVg is constant for an ideal gas where g = cP/cV, the ratio of molar specific heats. For an ideal gas the work W = (P1V1 - P2V2)/(g - 1)

### Isobaric Process

### Example

Gas Heated in a Cylinder fitted with a movable frictionless piston.
The pressure the atmosphere and the pressure due to the weight of the piston
remains constant as the gas heats up and expands.
First Law Implications: Δ*U* = *Q* - *W*
Unlike some of the other processes below neither the heat Q , the work W, or the change in internal energy Δ*U* are necessarily zero in a constant pressure process. For an ideal gas, constant pressure work is easily determined, *W* = *Ú PdV* = *P ΔV*
Part of the heat that flows into the system causes the temperature to rise, Q = n cp DT = m Cp DT, the rest goes into work.