Question

At constant pressure, the $\mu$ – T diagram for a pure substance that sublimes (s = solid, l = liquid and g = gas) is 

 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

In the $\mu$–T diagram for a substance that sublimes, the chemical potential of the solid phase decreases with temperature, while the chemical potential of the gas phase increases.

Answer 1

The Correct Option is C

$\mu-T$ Diagram Analysis

The stable phase of a substance at a given temperature ($T$) and pressure ($P$) is the one with the lowest chemical potential ($\mu$).

The relationship between the chemical potential ($\mu$) and temperature ($T$) at constant pressure is given by the derivative of the Gibbs free energy ($G$) with respect to $T$:

$$\left(\frac{\partial \mu}{\partial T}\right)_P = \left(\frac{\partial G/n}{\partial T}\right)_P = -\frac{S}{n} = -\bar{S}$$

where $\bar{S}$ is the molar entropy, which is always positive ($\bar{S} > 0$).

Therefore, the slope of the $\mu-T$ curve for any phase is negative:

$$\text{Slope} = \left(\frac{\partial \mu}{\partial T}\right)_P = -\bar{S} < 0$$

1. Slope Order

Since the slope is equal to the negative of the molar entropy, the magnitude of the slope is proportional to the molar entropy ($\text{slope magnitude} \propto \bar{S}$). Entropy increases with disorder. The molar entropies typically follow the order:

$$\bar{S}_{\text{gas}} (\text{g}) > \bar{S}_{\text{liquid}} (\text{l}) > \bar{S}_{\text{solid}} (\text{s})$$

This means the slopes of the $\mu-T$ curves will follow the order:

$$\text{Slope}_{\text{gas}} < \text{Slope}_{\text{liquid}} < \text{Slope}_{\text{solid}}$$

(The gas phase curve has the steepest negative slope, and the solid phase curve has the least steep negative slope).

Solid (s): Smallest negative slope.

Liquid (l): Intermediate negative slope.

Gas (g): Largest negative slope (steepest).

2. Sublimation at Low Temperature

Sublimation is the transition directly from solid ($\text{s}$) to gas ($\text{g}$) without passing through the liquid phase. For a substance to sublime at the pressure of the diagram, the solid-liquid (s-l) equilibrium curve must be above the liquid-gas (l-g) curve. In other words, the solid-gas transition point must occur before the solid-liquid transition point.

In a $\mu-T$ diagram, sublimation occurs when the solid ($\mu_{\text{s}}$) and gas ($\mu_{\text{g}}$) curves intersect at the lowest temperature $T_{\text{sub}}$. At temperatures below $T_{\text{sub}}$, the solid phase must have the lowest chemical potential (i.e., $\mu_{\text{s}}$ is the lowest line).

Analysis of Option (C):

Slope Order: The slopes correctly follow the order of $\mu_{\text{g}}$ being the steepest negative slope, $\mu_{\text{l}}$ being the intermediate negative slope, and $\mu_{\text{s}}$ being the least negative slope.

Phase Stability:

At the lowest temperatures shown, $\mu_{\text{s}}$ is the lowest curve, indicating solid is stable.

As temperature increases, the $\mu_{\text{g}}$ and $\mu_{\text{s}}$ curves intersect first. At the intersection, $\mu_{\text{s}} = \mu_{\text{g}}$, marking the sublimation temperature ($T_{\text{sub}}$).

Above $T_{\text{sub}}$, the $\mu_{\text{g}}$ curve is lower than the $\mu_{\text{s}}$ curve, indicating gas is stable. The liquid curve is always above the stable phase boundary in this region, confirming the liquid phase is never stable at this pressure (the substance sublimes).

The diagram (C) is the only one that correctly shows the slope order ($\text{Slope}_{\text{g}} < \text{Slope}_{\text{l}} < \text{Slope}_{\text{s}}$) and the $\mu_{\text{s}}-\mu_{\text{g}}$ intersection occurring at the lowest point, making solid-gas the favored transition (sublimation).