# Interpreting straight lines in a graph of isotherms of carbon dioxide

I am having difficulty in understanding pressure-volume graphs of $\ce{CO2}$ at different temperatures which describes liquefaction of $\ce{CO2}$.

I know that at the critical temperature of $\pu{31.1^\circ C}$ it is possible to liquefy the gas at the critical pressure of $\pu{73atm}$.

But I am having difficulty in understanding some statements about the graph.

(i)At temperatures below $\pu{30.98^\circ C}$ the behaviour of $\ce{CO2}$ towards compression is quite different.For example at $\pu{21.5^\circ C}$, $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point $\mathrm B$ a liquid of certain volume appears.Further compression do not change pressure but volume decreases till point $\mathrm C$ is reached. In fact, the gas and liquid $\ce{CO2}$ coexist along the horizontal line BC.The decreases of volume represents condensation of more and more $\ce{CO2}$ gas till the point C is reached.

I know that if we are compressing a gas its pressure should increase but why its pressure is constant along the line $\mathrm{BC}$?

(ii)At point C all the gases has been condensed and further application of pressure simply compresses the liquid .The line $\mathrm{CD}$ represents the compressibility of liquid Carbon Dioxide.

I have studied in my earlier classes that liquids cannot be compressed but how can the liquid $\ce{CO2}$ has a tendency of compression here?

• (1) As we compress the gas along the line BC, some of it turns to liquid, and we're left with a smaller amount of gas which exerts the same pressure. (2) True, liquids cannot be compressed... unless you press really hard. – Ivan Neretin Sep 19 '18 at 11:17
• Why should it exert same pressure as smaller amount of gas molecules means less collisions with the walls of container so it should have lower pressure. – pranjal verma Sep 19 '18 at 11:28
• You said if we are compressing a gas its pressure should increase. Why? – Ivan Neretin Sep 19 '18 at 11:52
• The blue area is a mixture of gas and liquid, and Ivan Neretin has explained what happens here. The area roughly V3 to H to E to V3 is liquid, the rest gas. The graph is poorly drawn and the line C-D, and similar, ones should be almost exactly vertical because as you mention liquids are effectively incompressible at the sort of pressures used here. – porphyrin Sep 19 '18 at 11:55
• The graph is good enough for our purposes. (Also, it is repeated in pretty much every book on the subject.) Now to the point. The pressure decreases because there are fewer molecules, and at the same time increases because old man Boyle said so. As a result, it stays the same. – Ivan Neretin Sep 19 '18 at 12:25

That these processes occur just enough to maintain line segment BC as an isobar can be understood more quantitatively from thermodynamics. Phase coexistence actually places significant constraints on the system, as is quantified by Gibbs' phase rule---a system of a single chemical species at two-phase coexistence is specified by a single parameter, here the temperature $T$, because we are considering an isotherm. This implies that the pressure is known given $T$; that is, we have a relationship $P = P(T)$, and of course this is just the liquid-vapor coexistence line on a phase diagram. This pressure is, in particular, independent of the number of moles of substance in each phase, and so we must have the same pressure at phase coexistence; i.e., over the line segment BC.