Question

Write the Van der Waals equation for one mole of gas.

Answer 1

Think about the Van der Waals equation that is defined and in which parameters of the ideal gas equation, the changes have been brought about to make it suitable for a real gas. Consider the constants used to do so.

Complete step by step solution:
We know that the Van der Waals equation has been derived due to the fallacies seen in the ideal gas equation. No gas can behave ideally under all conditions, there are some that come close, but some deviate a lot from the trend that should be seen. To combat this, the Van der Waal equation has been formulated that takes into consideration some real-life scenarios and modifies the values of pressure and volume. The values of the number of moles of substance present and its temperature are kept the same. The real gas equation is defined as:
$\left( P+\dfrac{{{n}^{2}}a}{{{V}^{2}}} \right)\left( V-nb \right)=nRT$
Here, $P$ are the ambient temperature that is exerted on the gas, $n$ is the number of moles of gas that are present, $V$ is the volume of the gas that is present. $a$ and $b$ are the constants that are substance-specific and used to define the real gas equation.
Now, the question asks us for the Van der Waal equation for 1 mole of a gas. So we will substitute the value of $n$ as 1. From this, we get:

& \left( P+\dfrac{{{1}^{2}}\times a}{{{V}^{2}}} \right)\left( V-1\times b \right)=1\times RT \\\ & \left( P+\dfrac{a}{{{V}_{m}}^{2}} \right)\left( {{V}_{m}}-b \right)=RT \\\ \end{aligned}$$ Here, ${{V}_{m}}$ is defined as the volume per mole of a substance. It can also be called the molar volume. Its value at STP (standard temperature pressure) will be $22.4L$. But here, we do not know what the temperature is going to be, so the volume cannot be predetermined. **Note:** Note that although the values of the Van der Waals constant change according to the substance that we are concerned with, they have definite units. $a$ has the unit ${{L}^{2}}bar/mo{{l}^{2}}$ and the value for $b$ has the unit $L/mol$.