# Chris Fox's Engineering Section

## Ideal Gases

*"For a gas at constant temperature, the pressure is inversely proportional to the volume."*

Boyle's Law

*"For a gas at constant pressure, the volume is directly proportional to the absolute temperature."*

Charles' Law

*"For a gas at constant volume, the pressure is directly proportional to the absolute temperature."*

Pressure Law

The above laws are brought together in the ideal gas equation pV=nRT. n is the number of moles, a measure of the quantity of matter in a gas; and R is the Universal Molar Gas Constant, whose value is 8.31 J mol^{-1} K^{-1}. The ideal gas is a good approximation for many gases.

For a unit mass of ideal gas, the change in heat energy is given by ΔQ=cΔθ (or cΔT), in which case ΔQ∝ΔT.

Consider constant volume and constant pressure processes. For a gas whose volume is constant, δW=pdV=0. Substituting into the First Law of Thermodynamics equation gives ΔQ=dU. Also, ΔQ≡ΔT. For a constant volume process, ΔQ=c_{v}ΔT. Therefore, dU=c_{v}ΔT. c_{v} is the specific heat capacity at constant volume, and it can be measured.

For a constant pressure process, ΔQ_{2}≡ΔT, where 2 is an arbitrary subscript. Therefore, ΔQ_{2}=c_{p}ΔT, where c_{p} is the specific heat capacity at constant pressure and ΔQ_{2}≡δQ. Substituting for the various parts of the First Law equation gives c_{p}ΔT-pΔV=c_{v}ΔT. At constant pressure, pΔV=nRΔT. Therefore, c_{p}ΔT-nRΔT=c_{v}ΔT. Dividing both sides by ΔT and rearranging gives c_{p}-c_{v}=nR.

An adiabatic process is one in which δQ=0. Many real life situations can be approximated to an adiabatic process. This time, substituting for the various parts of the First Law equation gives -pdV=c_{v}dT. Differentiating the ideal gas equation with respect to T gives V(dp/dT)+p(dV/dT)=nR (If two quantities are equal, their derivatives with respect to the same variable will also be equal). Rearranging this gives (V/nR)dp+(p/nR)dV=ΔT. Because -pdV=c_{v}dT, pΔV+c_{v}ΔT=0. Substituting for ΔT gives pΔV+c_{v}{(v/nR)Δp+(p/nR)ΔV}=0. Multiplying throughout by nR and rearranging gives pΔV(nR+c_{v})+c_{v}VΔp=0, and substituting for nR+c_{v} gives c_{p}pΔV+c_{v}VΔp=0. Dividing throughout by c_{p}pV gives γΔV/V+Δp/p=0, where γ=c_{p}/c_{v}. Now let us integrate. γ∫(1/V)ΔV+∫(1/p)Δp=0 leads to γ ln V + ln p = k, where k denotes a constant. Taking e to the power of both sides and simplifying gives pV^{γ}=k (e^{k} gives another constant).

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