Answer to Question #88750 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Calculate van der Waals' constants for
helium using the data Tc = 5.3 K,
pc = 2.25 atm and R = 8.31 J morl K-1.

Answer 1

The van der Waals' constants a and b are related to the critical temperature ${{T}_{c}}$ and pressure ${{p}_{c}}$ by

$$a=\frac{27T_{c}^{2}{{R}^{2}}}{64{{p}_{c}}},\,\,\,\,b=\frac{{{T}_{c}}R}{8{{p}_{c}}}$$

We have for helium ${{T}_{c}}=5.3\,K$ and ${{p}_{c}}\text{ }=\text{ }2.25\text{ }atm.$ Convert pressure in atmospheres to pressure in pascals:

$${{p}_{c}}=\text{1}\text{.01325}\cdot \text{1}{{\text{0}}^{5}}\,\left( Pa/1\,atm \right)\cdot 2.25\text{ }atm\approx \text{2}\text{.28}\cdot \text{1}{{\text{0}}^{5}}\text{ }Pa$$

Substituting the known values into these relations, we get

$$a=\frac{27\cdot {{\left( 5.3\,K \right)}^{2}}{{\left( 8.31\text{ }J\cdot mo{{l}^{-1}}\,\cdot {{K}^{-1}} \right)}^{2}}}{64\cdot \text{2}\text{.28}\cdot \text{1}{{\text{0}}^{5}}\text{ }Pa}$$

$$\approx 0.0036\,\frac{{{J}^{2}}}{Pa\cdot mo{{l}^{2}}}=0.0036\,\frac{Pa\cdot {{m}^{6}}}{mo{{l}^{2}}}$$

$$b=\frac{5.3\,K\cdot 8.31\text{ }J\cdot mo{{l}^{-1}}\,\cdot {{K}^{-1}}}{8\cdot \text{2}\text{.28}\cdot \text{1}{{\text{0}}^{5}}\text{ }Pa}\approx \text{2}\text{.4}\cdot \text{1}{{\text{0}}^{-5}}\frac{J}{Pa\cdot mol}=\text{2}\text{.4}\cdot \text{1}{{\text{0}}^{-5}}\frac{{{m}^{3}}}{mol}$$

To convert a from $Pa\text{ }\cdot {{m}^{6}}/mo{{l}^{2}}$ to $Pa\cdot {{L}^{2}}/mo{{l}^{2}}$ multiply by 10⁶

$$a=3.6\,\cdot {{10}^{3}}\frac{Pa\cdot {{L}^{2}}}{mo{{l}^{2}}}$$

To convert b from ${{m}^{3}}/mo{{l}}$ to $L/mo{{l}}$ multiply by 10³

$$b=\text{2}\text{.4}\cdot \text{1}{{\text{0}}^{-2}}\frac{L}{mol}$$

To convert a from $Pa\cdot \text{ }{{L}^{2}}/mo{{l}^{2}}$ , to $bar\cdot \text{ }{{L}^{2}}/mo{{l}^{2}}$ multiply by 10^-5

$$a=3.6\,\cdot {{10}^{-2}}\frac{bar\cdot {{L}^{2}}}{mo{{l}^{2}}}$$

Thus

$$a=0.0036\,\frac{Pa\cdot {{m}^{6}}}{mo{{l}^{2}}}=3.6\,\cdot {{10}^{3}}\frac{Pa\cdot {{L}^{2}}}{mo{{l}^{2}}}=3.6\,\cdot {{10}^{-2}}\frac{bar\cdot {{L}^{2}}}{mo{{l}^{2}}}$$

$$b=\text{2}\text{.4}\cdot \text{1}{{\text{0}}^{-5}}\frac{{{m}^{3}}}{mol}=\text{2}\text{.4}\cdot \text{1}{{\text{0}}^{-2}}\frac{L}{mol}$$