Switching to logarithmic coordinates, the reversible curves turn into straight lines. The trick is to take the logarithm of the equations of each process: powers become slopes, and every transformation acquires a gradient that identifies it at a glance.

Let us start from the equations of each process:

  • Isochore: V=constlnV=constV = \text{const} \Rightarrow \ln V = \text{const}, a vertical line;
  • Isobar: P=constlnP=constP = \text{const} \Rightarrow \ln P = \text{const}, a horizontal line;
  • Isotherm: PV=nRT=constlnP+lnV=constPV = nRT = \text{const} \Rightarrow \ln P + \ln V = \text{const}, a line of slope 1-1;
  • Adiabat: PVγ=constlnP+γlnV=constPV^\gamma = \text{const} \Rightarrow \ln P + \gamma\ln V = \text{const}, a line of slope γ-\gamma.

Since γ>1\gamma > 1, the adiabat is steeper than the isotherm even in these coordinates: its slope γ-\gamma is more negative than 1-1. The four transformations are thus immediately distinguished by their gradient.

lnP\ln PlnV\ln V diagram: the curves become straight lines. The adiabat has slope γ-\gamma (steeper), the isotherm has slope 1-1, the isobar is horizontal and the isochore is vertical.

Slope of the adiabat

From ΔSgas=0\Delta S_\text{gas} = 0 and T=PV/(nR)T = PV/(nR) one obtains the relation between the initial state AA and the final state BB: lnPBPA+γlnVBVA=0\ln\frac{P_B}{P_A} + \gamma\ln\frac{V_B}{V_A} = 0 Setting Y=lnPY = \ln P and X=lnVX = \ln V this becomes YBYA=γ(XBXA)Y_B - Y_A = -\gamma(X_B - X_A), that is, a line of slope γ-\gamma.

Area and work

In the PP-VV diagram the area under the curve represents the work done by the gas. In the lnP\ln P-lnV\ln V diagram the area no longer has this meaning: the change of coordinates distorts the areas. The logarithmic diagram is useful for visualising the curves more clearly (straight lines with distinctive slopes), but to calculate the work one must go back to the PP-VV diagram or use the formulas.

Topics: Thermodynamics Concepts: Thermodynamic transformations · Ideal gas law Skills: Reading graphs Methods: P-V and ln P - ln V diagram

Related exercises: Problem — Rectangular cycle in the p-V plane · Problem — Adiabatic compression of a diatomic gas · Problem — Stirling engine