# Isothermal and adiabatic compression of an ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.35 – 1.36.

When we compress an ideal gas, we can do so in a variety of ways. If we do the compression very slowly so that the gas is always in thermal equilibrium with its environment, then the temperature stays constant and the compression is isothermal.

On the other hand, if we compress the gas quickly so that it doesn’t have a chance to exchange heat with its environment, the temperature will change. This sort of compression is called adiabatic. [Incidentally, this use of the term ‘adiabatic’ has nothing to do with an adiabatic process in quantum mechanics and, in fact, the meaning is almost the opposite. In thermodynamics, an adiabatic compression occurs quickly so that heat doesn’t have a chance to escape; in quantum mechanics, an adiabatic process is one in which the Hamiltonian changes slowly. I’m not sure why the term is used in quantum mechanics as it’s derived the Greek adiabatos, which means ‘impassable’, so its use in thermodynamics as a process that is impassable to heat makes sense.]

Schroeder works out the details for these two types of compression in his section 1.5, so I’ll just review the main ideas here.

Isothermal compression

For isothermal compression, the temperature is constant so from the ideal gas law

$\displaystyle P=\frac{NkT}{V}=\frac{\mbox{constant}}{V} \ \ \ \ \ (1)$

The plot of ${P}$ versus ${V}$ on a PV diagram is a hyperbola known as an isotherm.

Since the thermal energy of a gas whose molecules have ${f}$ degrees of freedom is given by

$\displaystyle U=\frac{f}{2}NkT \ \ \ \ \ (2)$

the thermal energy of an isothermal process doesn’t change, so the work required to compress the gas must be entirely converted into the heat lost during the isothermal process. If the gas is compressed from volume ${V_{i}}$ to ${V_{f}}$, the work and heat are

$\displaystyle Q=-W=NkT\ln\frac{V_{f}}{V_{i}} \ \ \ \ \ (3)$

If the gas is being compressed, ${V_{f}, ${Q<0}$ (meaning that the gas loses heat) and ${W>0}$ (work is done on the gas).

Even though adiabatic compression must be done quickly to prevent heat loss, it can still be done slowly compared to the speed of sound in the gas, so that the quasistatic assumption (that the pressure of the gas is always uniform throughout the volume) is valid. For example, if you pump up a bicycle tire, you’ll find that the pump gets hot, but the speed at which you push the pump is significantly less than the speed of sound, so the compression is, to a good approximation, quasistatic.

By combining the equipartition theorem with the ideal gas law (details in Schroeder section 1.5) we arrive at a differential equation relating ${T}$ and ${V}$:

$\displaystyle \frac{f}{2}\frac{dT}{T}=-\frac{dV}{V} \ \ \ \ \ (4)$

At a given temperature and volume, if ${V}$ is decreased (the gas is compressed), the temperature rises. Integrating this equation gives the relation

$\displaystyle VT^{f/2}=K \ \ \ \ \ (5)$

where ${K}$ is a constant.

We can get a relation between ${P}$ and ${V}$ by using the ideal gas law in this equation.

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{PV}{Nk}\ \ \ \ \ (6)$ $\displaystyle VT^{f/2}$ $\displaystyle =$ $\displaystyle \frac{V^{1+f/2}P^{f/2}}{\left(Nk\right)^{f/2}}=K\ \ \ \ \ (7)$ $\displaystyle V^{\left(2+f\right)/2}P^{f/2}$ $\displaystyle =$ $\displaystyle K\left(Nk\right)^{f/2}\ \ \ \ \ (8)$ $\displaystyle V^{\left(2+f\right)/f}P$ $\displaystyle =$ $\displaystyle K^{2/f}Nk \ \ \ \ \ (9)$

or, defining the adiabatic exponent ${\gamma\equiv\left(2+f\right)/f}$, we have the relation for a quasistatic adiabatic process:

$\displaystyle \boxed{V^{\gamma}P=\mbox{constant}} \ \ \ \ \ (10)$

Example Suppose we compress a litre of air (assumed to be an ideal gas of diatomic molecules, with 5 degrees of freedom) at atmospheric pressure to a pressure of 7 atm. If the process is adiabatic, then the final volume is found from

 $\displaystyle \gamma$ $\displaystyle =$ $\displaystyle \frac{7}{5}\ \ \ \ \ (11)$ $\displaystyle P_{i}V_{i}^{7/5}$ $\displaystyle =$ $\displaystyle 1\mbox{ atm litre}^{7/5}\ \ \ \ \ (12)$ $\displaystyle P_{f}V_{f}^{7/5}$ $\displaystyle =$ $\displaystyle 7V_{f}^{7/5}\ \ \ \ \ (13)$ $\displaystyle V_{f}$ $\displaystyle =$ $\displaystyle \frac{1}{7^{5/7}}=0.249\mbox{ litre} \ \ \ \ \ (14)$

The work done in the compression is

 $\displaystyle W$ $\displaystyle =$ $\displaystyle -\int_{V_{i}}^{V_{f}}P\left(V\right)dV\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{1}^{0.249}\frac{dV}{V^{7/5}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5}{2}\left.\frac{1}{V^{2/5}}\right|_{1}^{0.249}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.86\mbox{ atm litre}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.86\times1.013\times10^{5}\times10^{-3}\mbox{ J}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 188.4\mbox{ J} \ \ \ \ \ (20)$

From 5 we can find the temperature after compression, given that the initial temperature is ${T_{i}=300\mbox{ K}}$. We have

 $\displaystyle \frac{V_{f}T_{f}^{5/2}}{V_{i}T_{i}^{5/2}}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (21)$ $\displaystyle T_{f}$ $\displaystyle =$ $\displaystyle \left(\frac{V_{i}}{V_{f}}\right)^{2/5}T_{i}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{1}{0.249}\right)^{2/5}300\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 523\mbox{ K} \ \ \ \ \ (24)$

That is very hot; it’s approaching the melting point of lead (600 K). This pressure is actually in the range of a road bicycle tire, so it’s not surprising that a bicycle pump can get very hot when you’re inflating a tire.

## 19 thoughts on “Isothermal and adiabatic compression of an ideal gas”

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4. Benjo

Hi, first of all thank you for your carefully written solutions! It is a blessing for me.

I think in equation (16) you are missing the constant C=p*V^γ which you had to evaluate from one pair of p and V before. Taken into account this constant C in the calculation, you get a slightly higher result of 204.3 J.

Thanks again and keep going please!

1. gwrowe Post author

When working out eqn 16 I am using units of atm-litre, and from eqn 12, the constant is ${P_{i}V_{i}^{7/5}=1}$, so the value of 1.86 atm litre in eqn 18 is actually correct, and this comes out 188.4 J as stated in eqn 20. You’d only need to include a non-1 constant if you worked in joules from the start.

5. confused

how do you get to (20) from (18).
how can atm.litre go to Joules?

also. where does the “1” in (14) come from?

1. gwrowe Post author

atm-litre has the units of energy (pressure times volume) so can be converted to joules.
The initial pressure is P_i = 1 atm, hence the ‘1’ in (14).