# Elaboration of Boyles Law

1. Apr 5, 2013

### Aj83

As the Boyle's law states that there is an inverse relationship between pressure and volume at constant temperature, so why does the temperature increases in practice when for example air is compressed i,e increase in pressure and decrease in volume?

2. Apr 5, 2013

### Staff: Mentor

To put it simply, because you are putting more energy in the gas when you compress it. In other words, yes, P increases and V decreases, but P increases more than V decreases, so T must increase. If you want to compress a gas without changing its temperature, you have to do it slowly enough such that heat can flow out of the gas at the same rate as you are putting it in.

3. Apr 5, 2013

### Aj83

So does that mean that ideal gas equation which is based on boyle's law is not applicable unless the expansion or compression is isothermal?

4. Apr 5, 2013

### Staff: Mentor

It is still applicable. Indeed, it is the basis for the explanation I gave you.

[STRIKE]Boyle's law is simply $PV \propto T$, which the ideal gas law expands by setting the proportionality constant,[/STRIKE]
Boyle's law is simply $PV = \text{const.}$ (for a given amount of gas at constant $T$), so it is "included" in the ideal gas law:
$$PV = N k_B T = n R T$$

If you compress a gas, the work done on the gas is
$$W = -\int_{V_i}^{V_f} P dV$$
If $P$ is constant, you then have $W = -P \Delta V$, otherwise you need to rewrite $P$ as a function of $V$ (for instance, by using the ideal gas law). At the same time, you have the change in energy of the gas as
$$dU = Q + W$$
where $Q$ is the heat entering the gas. If $Q=0$ (adiabatic process), then $dU = W$, which is greater than 0 if you are compressing the gas. Considering that the change of energy for an ideal gas is also given by
$$dU = N \frac{f}{2} k_B dT$$
where $f$ is the number of degrees of freedom of the gas ($f=3$ for a monatomic gas), you see that $T$ must increase if $U$ increases. The only way around that is to have $Q=-W$: you have to take out heat from the gas at the rate you are putting it in by compressing.

Last edited: Apr 5, 2013
5. Apr 5, 2013

### Aj83

Thanks for your replies, I understand why the temperature increases when the work is done on the gas (in adiabatic compression work input goes on to increase the internal energy hence the increase in temperature). I was just trying to understand it from the Boyle's law point of view and it's usefulness.

6. Apr 5, 2013

### technician

Boyle's law is simply PV∝T, which the ideal gas law expands by setting the proportionality constant,

I don't think this is Boyle's law !!!!!! (where have you seen it quoted like this??)
Boyle's law is that PV = constant as long as the temperature is kept constant.

7. Apr 5, 2013

### Aj83

Technician you are right, I noticed that too but didnt correct it. I think the equation DrClaude wrote is when Boyle's and Charle's Laws are combined to derive ideal gas equation.

8. Apr 5, 2013

### Staff: Mentor

You are absolutely right. I'm sorry for the mistake, I was doing it from the top of my head.

9. Apr 5, 2013

### technician

Thanks for clearing it up Dr

10. Apr 5, 2013

### Staff: Mentor

No. It means that Boyle's Law which is based on the ideal gas equation is not applicable unless the the expansion or compression is isothermal. The ideal gas equation is more general.