## Changing the Melting Point

### Task number: 2178

Molar entropy change of melting ice is 22.2 J mol^{−1}K^{−1}. Determine the change in the melting point if the external pressure has increased from 100 kPa to 200 kPa.

#### Hint 1

To calculate the temperature change Δ

*T*, use the so-called Clapeyron equation. Determine which quantities in the equation we know and which ones we have to express.#### Hint 2

Since the temperature lies in a relatively small range, the molar entropy change Δ

*S*_{mol}, and hence the whole right side of the Clapeyron equation, can be considered constant.#### Notation

Δ *S*_{mol}= 22.2 J mol^{−1}K^{−1}molar entropy change of melting ice *p*_{1}= 100 kPa = 100 000 Painitial pressure *p*_{2}= 200 kPa = 200 000 Pafinal pressure Δ *T*= ?change in melting point

*From the Tables:**ρ*_{w}= 1000 kg m^{−3}density of water *ρ*_{i}= 917 kg m^{−3}density of ice *M*_{m}= 0.018 kg mol^{−1}molar mass of water #### Analysis

The phase transition point depends on pressure. Typical example is a pressure cooker, in which water is boiling at a temperature considerably higher than 100 °C. On the contrary, in high mountains, the boiling point of water is decreased because of the lower pressure.

These changes can be expressed quantitatively using the so-called Clapeyron equation. Basically, it is a differential equation stating that the ratio of the pressure change and phase transition temperature change is equal to the ratio of the molar heat of the transition and the product of the thermodynamic temperature and the difference between the molar volume of the substance in the 2

^{nd}and in the 1^{st}phase.In order to obtain a formula consisting of the quantities given in the task only, we need to adjust the equation. First, we can substitute the ratio of the molar heat and the temperature by the molar entropy change. Second, we express the molar volume, both in the 1

^{st}and in the 2^{nd}phase, as the ratio of the molar mass and corresponding density. After we express the pressure change as the difference between the final and initial pressure, we obtain a relation from which we just need to express the temperature change.#### Solution

Our calculation is going to be based on the so-called

\[\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{L_{mol}}{T(V_{2m}-V_{1m})},\]*Clapeyron equation*, whose most common form is as followswhere

*p*stands for pressure,*T*for thermodynamic temperature at which the phase transition occurs,*L*_{mol}for molar heat and*V*_{1m}and*V*_{2m}for molar volume of the substance in the 1^{st}and 2^{nd}phase respectively.In the formula, molar heat

\[\mathrm{\Delta}S_{mol}=\frac{L_{mol}}{T},\]*L*_{mol}is unknown. However, remembering that the following relation applies to the molar entropy change Δ*S*_{mol}we can adjust the original formula as follows

\[\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{\Delta}S_{mol}}{V_{2m}- V_{1m}}.\]Then, we also need to express the molar volume of the ice

\[V_{1m}=\frac{M_{m}}{\rho_{i}},\] \[V_{2m}=\frac{M_{m}}{\rho_{w}},\]*V*_{1m}and of the water*V*_{2m}using the given quantities. We use the known relationwhere

*M*_{m}is the molar mass of water (as well as ice, of course),*ρ*_{i}density of ice and*ρ*_{w}density of water.Substituting the expressed quantities in the Clapeyron equation, we obtain

\[\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{\Delta}S_{mol}}{\frac{M_{m}}{\rho_{w}}-{\frac{M_{m}}{\rho_{i}}}}.\]Since the temperature lies in a relatively small range, the molar entropy change Δ

\[\frac{\mathrm{\Delta}p}{\mathrm{\Delta}T}=\frac{\mathrm{\Delta}S_{mol}}{\frac{M_{m}}{\rho_{w}}-{\frac{M_{m}}{\rho_{i}}}}.\]*S*_{mol}, and hence the whole right side of the Clapeyron equation, can be considered constant. Thanks to that, we do not have to solve the differential equation and we can just substitute the differential increments for classic “delta”:Therefrom, we can express the required temperature change Δ

\[\mathrm{\Delta}T=\frac{\mathrm{\Delta}p\left(\frac{M_{m}}{\rho_{v}}-{\frac{M_{m}}{\rho_{l}}}\right)}{\mathrm{\Delta}S_{mol}}.\]*T*right away:Finally, we express the pressure change Δ

\[\mathrm{\Delta}T=\frac {(p_{2}-p_{1}) \left(\frac{M_{m}}{\rho_{w}}-{\frac{M_{m}}{\rho_{i}} }\right)}{\mathrm{\Delta}S_{mol}}.\]*p*as the difference between the final pressure*p*_{2}and the initial one*p*_{1}#### Numerical Substitution

\[\mathrm{\Delta}T=\frac {(p_{2}-p_{1}) \left(\frac{M_{m}}{\rho_{w}}-{\frac{M_{m}}{\rho_{i}} }\right)}{\mathrm{\Delta}S_{mol}}=\frac {(200\,000-100\,000)\cdot \left(\frac{0.018}{1000}-\frac{0.018}{917}\right)}{22.2}\,\mathrm{K}\] \[\mathrm{\Delta}T\dot{=}-0.0073\,\mathrm{K}\]#### Answer

The melting point of ice decreased by approximately 0.0073 K.

#### Comment

When using the Clapeyron equation, we need to pay extra attention to the sign.

In our case, water has a higher density than ice, so its molar volume is smaller (

*V*_{2m}<*V*_{1m}). Thus, if the pressure increases (d*p*> 0), the phase transition point will logically decrease (d*T*< 0).Similar considerations must be done for every task of this type, otherwise the result will be numerically correct, but physically completely absurd!