 # 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=cvΔT. Therefore, dU=cvΔT. cv is the specific heat capacity at constant volume, and it can be measured.

For a constant pressure process, ΔQ2≡ΔT, where 2 is an arbitrary subscript. Therefore, ΔQ2=cpΔT, where cp is the specific heat capacity at constant pressure and ΔQ2≡δQ. Substituting for the various parts of the First Law equation gives cpΔT-pΔV=cvΔT. At constant pressure, pΔV=nRΔT. Therefore, cpΔT-nRΔT=cvΔT. Dividing both sides by ΔT and rearranging gives cp-cv=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=cvdT. 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=cvdT, pΔV+cvΔT=0. Substituting for ΔT gives pΔV+cv{(v/nR)Δp+(p/nR)ΔV}=0. Multiplying throughout by nR and rearranging gives pΔV(nR+cv)+cvVΔp=0, and substituting for nR+cv gives cppΔV+cvVΔp=0. Dividing throughout by cppV gives γΔV/V+Δp/p=0, where γ=cp/cv. 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 pVγ=k (ek gives another constant).