For a pure species, the slope of the melting line $\dfrac{dP}{dT}$ at $-2^{\circ}$C is $-5.0665\times 10^{6}$ Pa K$^{-1}$. The difference between the molar volumes of the liquid and solid phase at $-2^{\circ}$C is $-4.5\times 10^{-6}$ m$^{3}$ mol$^{-1}$. The value of the latent heat of fusion (rounded off to nearest integer) is _________ J mol$^{-1}$.
Show Hint
For phase transitions at equilibrium, the Clapeyron equation $\,\frac{dP}{dT} = \frac{L}{T\Delta V}\,$ directly connects pressure–temperature slope with latent heat. Sign conventions matter but the magnitude of $L$ is obtained from absolute values.
To find the latent heat of fusion, we use the Clapeyron equation that relates pressure, temperature, volume change, and latent heat across a phase transition:
\[
\frac{dP}{dT} = \frac{L}{T \Delta V}
\]
Step 1: Identify the given values:
Slope of melting line:
\[
\frac{dP}{dT} = -5.0665 \times 10^6 \ \text{Pa K}^{-1}
\]
Difference in molar volumes (liquid – solid):
\[
\Delta V = -4.5 \times 10^{-6} \ \text{m}^3 \text{ mol}^{-1}
\]
Temperature:
\[
T = -2^{\circ}\text{C} = 271 \ \text{K}
\]
Step 2: Apply the Clapeyron equation to solve for $L$:
\[
L = \frac{dP}{dT} \times T \times \Delta V
\]
Substitute the values:
\[
L = (-5.0665 \times 10^{6}) \times (271) \times (-4.5 \times 10^{-6})
\]
Step 3: Compute the product step-by-step:
First multiply the slope and the volume change:
\[
(-5.0665 \times 10^{6}) \times (-4.5 \times 10^{-6})
= 5.0665 \times 4.5
= 22.79925
\]
Now multiply by temperature:
\[
L = 22.79925 \times 271 = 6188.65
\]
Step 4: Round off to nearest integer:
\[
L \approx 6189 \ \text{J mol}^{-1}
\]
Thus, the latent heat of fusion is approximately 6189 J mol\(^{-1}\).
