# Gaseous States and Laws [closed]

What are the factors on which the value of the Boyle temperature depends? And if the Boyle temperature depends on the pressure, then why?

As at the Boyle temperature a real gas becomes ideal, it should be independent of pressure, but on reading an article on internet it confuses me so please help me understand why.

## closed as off-topic by A.K., Tyberius, aventurin, Mithoron, Todd MinehardtNov 22 '18 at 17:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• The energy of an ideal gas is independent of pressure, but what do you mean by "it should be independent of pressure". – Buck Thorn Nov 21 '18 at 20:19

It is defined as the temperature at which the second virial coefficient $$B_2(T)$$ becomes zero. Near that temperature, over a wide range of pressures, it is true that the equation of state of the gas is quite close to the ideal gas equation, because typically the higher virial coefficients give a relatively small contribution. However, it is not quite right to say that the gas "becomes" an ideal gas. The behaviour is really a consequence of two cancelling contributions, at that particular temperature.
You can see a formula for the dependence of $$B_2$$ on both the temperature and the pair potential here. So, if you know the pair potential, and can do the integral (numerically, perhaps) you can calculate $$B_2(T)$$ for any temperature. If you plot it as a function of $$T$$, almost always it will go through zero somewhere: this defines the Boyle temperature.
The reason that there is a temperature at which $$B_2$$ becomes zero, for most gases, is that at very high temperatures $$B_2$$ is positive, being determined by the repulsive interactions between atoms (which reflect the fact that atoms have a nonzero size), while at low temperatures the attractive forces play the largest part in determining $$B_2$$, and these make it negative. So somewhere in between, it must go through zero. If you look on this page at the section on the (approximate) van der Waals equation of state, you can see these two contributions to $$B_2$$: $$b$$ (atomic size) and $$-a/RT$$ (attractive forces). So in this case, the Boyle temperature, where $$B_2$$ is zero, is $$T_\text{B}=a/(bR)$$.