# What is the interpretation of pressure times volume as energy?

I know that pressure times volume has units of energy, but is there an intuitive explanation of how the pressure contributes to the total energy? It seems clear for ideal gases using the $PV=nRT$ formalism, but is there a mechanistic explanation of why pressure embodies energy? I'm thinking of something along the lines of potential energy: "$U(x)$ is the energy required to move the point from $r=\infty$."

My first thought is to say that the pressure of an ideal gas in a compressed cylinder (with a frictionless piston on one end) tells us the total amount of work that the gas could perform on the piston if it were allowed to isothermally expand until $P=0$ (the piston is infinitely long):

$$P_0=\frac{nRT}{V_0}\implies \delta W=PdV=PA_{cyl}dx$$ (where $x$ is the longitudinal axis of the cylinder).

Under an isothermal process, $$PV=c \implies P=\frac{c}{V}$$ $$dV=A_{cyl}dx\implies W_{Total} = cA_{cyl}\int\limits_{x_0}^{\infty} \frac{1}{A_{cyl}x}dx=c \lim_{x\to \infty} \ln(x)-\ln(A_{cyl}x)=\infty$$ which is clearly wrong. However, I'm not sure what variables to integrate over to make it work out.

• I would say it's more a case that pressure represents energyin the system. Dec 31, 2014 at 20:53
• Did you mean to say that pressure * volume has units of energy? That's implicit in your own formula $\delta W=PdV$. Dec 31, 2014 at 21:06
• @Hypnosifl you're quite right..I've corrected.
– user31580
Dec 31, 2014 at 21:55
• @CarlWitthoft what about temperature? Isn't it also representing the energy in the system?
– user31580
Dec 31, 2014 at 21:56
• $PV = nRT$ often called the 'ideal gas law' is also called the 'state' equation, presumably since it relates the thermodynamic state of the gas to a definite amount of energy. What is it saying? The left hand side (energy) is proportional to the temperature of an 'n' amount of particles. Right? Dec 31, 2014 at 22:54

Suppose a gas is contained inside a container of total surface area $S$, which can somehow expand under the effect of the pressure. You can then compute the work done by this pressure by evaluating the force as $\text d\mathbf F = p\hat{\mathbf n}\text dS$ applied to an infinitesimal surface $\text dS$ around a generic point $\mathbf x$ on the surface $\Sigma$ of the container. If $\delta\mathbf x$ is an infinitesimal displacement field on $\Sigma$, the work done by the pressure is estimated by $$\delta W = \int_\Sigma p\delta\mathbf x\cdot\hat{\mathbf n}\text dS.$$ Further assume that it is known how the surface $\Sigma$ evolves in time, that is you have $\Sigma(t)$ and $\mathbf v$ is the velocity field with which every point of the surface moves. Then the total work would be $$W=\int_a^b\text dt\ p(t)\int_{\Sigma(t)}\mathbf v(\mathbf x,t)\cdot\hat{\mathbf n}\text dS$$ In the first formula, the product $\delta\mathbf x\cdot\hat{\mathbf n}\text dS$ can be interpreted as a $\delta V$, so that $\delta W = p\delta V$. The second formula, that for $W$, is more like the integral of a power over time, and of course "$\text dV = \mathbf v(\mathbf x,t)\cdot\hat{\mathbf n}\text dS\text dt$".
Think about the equivalence of energy with the product of pressure and volume this way: By the definition of work, $$W=F\cdot d$$. By the definition of pressure, $$P=\frac{F}{A}$$. Rearranging the second equation, $$F=PA$$. Substituting it into the first equation, $$W=PA\cdot d$$. Since $$Ad$$ is the volume, the net work done by the forces is $$W=PV$$. Since work is just energy transfer, $$E=PV$$, where E is energy, P is pressure, and V is volume.