Volume work

This article deals with closed systems, in particular a constant amount of substance (subject to chemical reactions); for flowing systems see hydraulics .

When a piston expands a distance against an external pressure , it does the volume work . ${\displaystyle \Delta z}$ ${\displaystyle p}$ ${\displaystyle W_{\rm {1,2}}}$

The volume of working or volume change work is in a closed system to be performed work , the volume of the system from the value of a with the value to alter: ${\displaystyle W}$ ${\displaystyle V_{1}}$ ${\displaystyle V_{2}}$

• in the case of volume reduction by compression , compression work is done, i. H. supplied to the system (in the figure this is the work that the piston does on the gas contained in the cylinder): ${\displaystyle (V_{2}<V_{1})}$ ${\displaystyle W>0}$
• in the case of volume increase through expansion, work - i. H. Energy  - free, d. H. issued by the system: ${\displaystyle (V_{2}>V_{1})}$ ${\displaystyle W<0.}$

The formula for volume work is:

${\displaystyle W_{1,2}=-\int \limits _{s}F(s)\cdot \mathrm {d} s}$.

Here is the force that acts along a path ; this is counted positively in the direction of expansion (in the illustration against the compression force shown ). ${\displaystyle F(s)}$ ${\displaystyle s}$ ${\displaystyle F_{p}}$

The minus sign in the formula is a convention ; This ensures that the work supplied to the system is positive, as described above, while the energy released is given a negative sign. In the compression shown, the distance covered has a negative sign which is compensated for by the additional minus sign in the formula for the volume work. ${\displaystyle \left(\mathrm {d} s<0\right),}$

Smooth process

The work supplied without friction and quasi-static is in the cylinder shown with the cross section ${\displaystyle A}$ ${\displaystyle \left(\Rightarrow \mathrm {d} s={\frac {\mathrm {d} V}{A}}\right)}$

because of (freedom from friction): ${\displaystyle F=p\cdot A}$

${\displaystyle \Rightarrow W_{\mathrm {1,2} }=\int \limits _{V_{1}}^{V_{2}}\delta W=-\int \limits _{V_{1}}^{V_{2}}p\cdot \mathrm {d} V}$

With

• ${\displaystyle \delta W=-pdV}$the inexact differential of volume work
• ${\displaystyle p}$: Pressure
• ${\displaystyle \mathrm {d} V}$: Volume change.

This change of state runs in the pV diagram from point 1 to point 2, with the compression shown, i.e. in the negative volume direction, the compression work would therefore have a negative sign without the minus sign in the formula. ${\displaystyle \left(\mathrm {d} V<0\right);}$

The integral value, which corresponds to the area under the state curve, can be calculated if the function  p = f (V) is known (see below).

Frictional process

In the real case, if between the piston and the cylinder, a frictional force acts must also when compressing the volume-changing work frictional work can be applied. This increases the internal energy of the system and thus the pressure compared to the frictionless process (if it is not dissipated to the outside as heat by cooling ): ${\displaystyle W_{R}}$

${\displaystyle {p_{2}}'>p_{2}}$

In the pV diagram, the change in state now runs from point 1 to point 2 '. This means that the work of volume change, which corresponds to the area under the curve, also increases without the work of friction itself being included :

${\displaystyle \Rightarrow W_{1,2'}>W_{1,2}}$

The work to be done from the outside is the sum of the now larger volume change work and the friction work:

${\displaystyle W_{\mathrm {ext} }=W_{1,2'}+W_{R}}$

Calculation example

Assume the isothermal expansion of an ideal gas ${\displaystyle \left(T={\text{konst.}}\right).}$

Then by inserting the thermal equation of state of ideal gases :

${\displaystyle p(V)=n\cdot R\cdot T\cdot {\frac {1}{V}}}$

With

• n is the amount of substance
• R is the general gas constant
• T is the absolute temperature

solve the integral for the volume work:

${\displaystyle {\begin{aligned}\Rightarrow W_{\mathrm {1,2} }&=-&n\cdot R\cdot T\cdot \ln {\frac {V_{2}}{V_{1}}}\\&=&n\cdot R\cdot T\cdot \ln {\frac {V_{1}}{V_{2}}}\end{aligned}}}$

This equation shows that when an ideal gas expands, the volume work is negative, i.e. energy is released; this follows from the logarithm , which is negative for numbers less than one and positive for numbers greater than one:

${\displaystyle {\begin{aligned}V_{2}>V_{1}\\\Leftrightarrow {\frac {V_{2}}{V_{1}}}>1\\\Leftrightarrow \ln {\frac {V_{2}}{V_{1}}}>0\\\Rightarrow W_{\mathrm {1,2} }<0\end{aligned}}}$

Instead of n R one can also use m R _s above :

${\displaystyle n\cdot R=m\cdot R_{\mathrm {s} }}$

in which

• m is the mass of the substance and
• R _{s is} its specific gas constant .

Open system

If the compression is carried out in an open system with the external pressure , then actual work must be done ${\displaystyle p_{0}}$

${\displaystyle W_{1,2}+p_{0}\cdot (V_{2}-V_{1})}$

applied, as the external pressure multiplied by the area also results in a force. If the external pressure is higher than the internal pressure of the volume to be compressed, then energy is gained; if it is less, work has to be done.

See also

• thermodynamics
• Dissipation
• Technical work
• Move work

literature

• Literature on technical thermodynamics