Discussion
m1 and m2 are the slopes (dP/dT) of the solid-liquid equilibrium lines in the P-T phase diagrams of H2O and CO2, respectively. For P \(\lt\) 10 atm, the values of m1 and m2 are
Answer: D
The slopes are \(m_{1} \lt 0\) and \(m_{2} \gt 0\).
The Clapeyron equation gives the slope of the solid-liquid line:
\[ \frac{dP}{dT} = \frac{\Delta H_{fus}}{T\,\Delta V_{fus}} \]
\(\Delta H_{fus}\) is always positive, so the sign of the slope is the sign of the volume change on melting. Ice is less dense than liquid water, so water contracts on melting (\(\Delta V_{fus} \lt 0\)) and its melting line slopes backwards: \(m_{1} \lt 0\). Solid CO2 is denser than the liquid, like most substances, so \(\Delta V_{fus} \gt 0\) and \(m_{2} \gt 0\).
Water is the unusual case, so any option giving both slopes the same sign, or reversing the two, is wrong.
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