The constant pressure condition

Contents

• The system under condition of
• The system under condition of p = const
• Physical justification
• The isothermal compressibility
• About this document ...

The system under condition of $p_{ex} =  const$

This is a very simple situation. All we are saying is that the external pressure, i.e. the pressure exerted by the surroundings on the system, is constant. We do not know anything about the system pressure p. It may vary or stay the same, depending on the details of the process in question.

The system under condition of p = const

Rather surprisingly, this is a quite complicated situation. Imagine that we have a glass filled with a water/ice mixture - as the system absorbs heat from the air in the room, the ice will melt and the volume occupied by the mixture will decrease, but the pressure will remain at it's normal value, about 1 atm. Another situation might be a test-tube with a reacting mixture under a hood - as the reaction progresses gaseous products may be released, the reacting mixture might heat up etc. Still, the pressure will remain at the normal value of 1 atm. How is this possible? The answer is that to assure conditions of constant pressure in our system, we have to put it in contact with a special kind of surroundings, which will adjust to changes in volume without changing their pressure (the technical name is a volume reservoir).

We prepare the surroundings in a state where the external pressure $p_{ex}$ is equal to our desired pressure. We place the system in contact with these surroundings. If the initial system pressure is higher than $p_{ex}$, the system will expand, if the initial system pressure is lower than $p_{ex}$, the system will be compressed. In the end, mechanical equilibrium will be reached:

$$p =  p_{ex}$$

Because of the special property of the surroundings, the changes in the system will not lead to a measurable change in the pressure in the surroundings, which will remain at its pre-determined value of $p_{ex}$. We can write this as

$$p =  p_{ex} =  const$$

This is the real meaning of the statement "the system is at a constant pressure p". Thus, a system at constant pressure p is actually in mechanical equilibrium with surroundings at the same pressure.

Physical justification

Let us think of the "supersystem" consisting of system + surroundings. In this isolated system, the volume is constant:

$$V_{tot} =  V +  V' =  const   or    \Delta V +  \Delta V' =  0$$

where V and V' are the volumes of the system and surroundings. This tells us that the volume changes in the system and surroundings are equal in magnitude but opposite in sign.

We will introduce the iosthermal compressibility of the surroundings

$$\kappa' =  -\frac{1}{V'}\left(\frac{\partial V'}{\partial p'}\right)_{T'}$$

where T' and p' refer to the temperature and pressure of the surroundings (p' here is the same as the $p_{ex}$ used in the previous sections; the "-" sign in the equation is there to make the quantity positive). The change in pressure $\Delta p'$ corresponding to a small change in volume $\Delta V'$ may then be calculated as

$$\Delta p ' \approx  -\frac{1}{\kappa'}\frac{\Delta V'}{V'}$$

How can we make $\Delta p'$ small? The easiest way is to make V' very large, much larger than the changes in system volume $\Delta V$ (which have the same magnitude as the changes in surroundings volume $\Delta V'$). Then the fraction $\Delta V'/V'$, i.e. the relative volume change in the surroundings, will be small. The atmosphere certainly has this property - the fractional volume change of the atmosphere caused by melting of ice in a glass is negligible. The atmosphere also has a relatively high compressibility $\kappa'$, another factor that makes $\Delta p'$ small.

Thus, the ideal volume reservoir has a large volume and large compressibility.

The isothermal compressibility

The isothermal compressibility

$$\kappa =  -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T}$$

is an important material property, characterizing the ease of changing the volume by applying pressure.

This quantity may easily be calculated for the ideal gas using the formula for p:

$$V =  \frac{nRT}{p}$$

Thus

$$\kappa =  -\frac{1}{V}\left(\frac{\partial (\frac{nRT}{p})}{\partial p} \right)_{T} =  \frac{nRT}{Vp^2} =  \frac{1}{p}$$

The ideal gas has higher compressibility at lower pressures.

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