Boyle's law- what's the big deal if temperature is not constant?

Question

Boyle's law: At constant temperature of the gas, the volume of a given mass of a gas is inversely proportional to its pressure.

So, Boyle's law is talking about isothermal condition,right? But, what if temperature is not constant?

My thinking: Suppose, there is a given gas in a frictionless cylinder fitted with a piston. Its initial temperature is $t_0$ and its pressure is $p_0$ . It is at equilibrium initially and hence the piston will also exert pressure $p_0$ so as to maintain mechanical equilibrium. Now heat is applied to it by a burner. As the gas gains thermal energy, the gas' temperature increases from $t_0$ to $t_k$ . Now as the molecules have more KE, the pressure exerted by the gas on the piston will be more now. And the gas will expand now as a result of which volume increases. But now, the pressure of the gas is low now as the KE is used to displace the piston. And so finally, we can write $$P\propto \frac{1}{v}$$ which is what Boyle's law says but temperature being same.

So, what is the importance of constant temperature?? The gas still follows the law even if temperature is variant. Or am I mistaking anywhere?? Please help. I am confused.

Answer 1

Suppose you compress a piston of gas, while decreasing its temperature drastically. You will find that as $V$ decreases, $P$ decreases as well, so $P$ is no longer proportional to $\frac{1}{V}$ .

Example: Initial state:$V = 1~\text{L}$, $T = 300~\text{K}$, $P = 1~\text{bar}$

Final state: $V = 0.5~\text{L}$, $T = 50~\text{K}$, $P = 0.3~\text{bar}$

Answer 2

I agree with answers of John Rennie and t.c, but would like to add a few examples.

• If the piston is displaced then work is done by the system (the system being the gas insider the cylinder), which is equal to $\int_{x_0}^{x_1} p_0 dx$ if the piston applies a constant pressure, $p_0$, and it starts at position $x_0$ and moves to position $x_1$. In this case the heat energy applied to the gas by the burner partly gets converted to work done in pushing the cylinder and partly heats up the gas - The final result would be gas at the same pressure $p_0$, but higher volume $V$ and higher temperature $T$ following the equation shown by John Rennie

In your question you talk about heating up the gas, the gas losing energy by doing work on the piston and its pressure dropping - for that case the piston must reduce its pressure on the gas, otherwise the piston would compress the gas.

The best example I can think of to relate directly to your question is if the cylinder is in a heat bath that keeps it at constant temperature and the piston is free to move. Then the volume, $V$, can increase and the pressure, $P$, will drop and heat will go into the gas, which will be converted into work done in pushing back the cylinder. The temperature of the gas will be constant.

Answer 3

The equation of state of an ideal gas is:

$$ PV = nRT $$

where $n$ is the number of moles of the gas (we often assume we're talking about one mole and omit $n$) and $R$ is the gas constant. Rearranging this gives:

$$ P = nRT\frac{1}{V} $$

So $P$ is only proportional to $V^{-1}$ if the constant of proportionality, $nRT$, is a constant i.e. only if $T$ is constant. If $T$ changes then $P$ is not proportional to $V^{-1}$.

Answer 4

The mathematical equation for an ideal gas undergoing a reversible (i.e., no entropy generation) adiabatic process is $P V^{\gamma} = \operatorname{constant} \qquad $

where P is pressure, V is volume, and

$ \gamma = {C_{P} \over C_{V}} = \frac{f + 2}{f}$, $C_{P}$ being the specific heat for constant pressure, $C_{V}$ being the specific heat for constant volume, $ \gamma $ is the adiabatic index, and $f$ is the number of degrees of freedom (3 for monatomic gas, 5 for diatomic gas and collinear molecules e.g. carbon dioxide). For a monatomic ideal gas, $\gamma = 5/3 $, and for a diatomic gas (such as nitrogen and oxygen, the main components of air) $\gamma = 7/5$.