## Ideal Gas Versus Van der Waals

### Task number: 1284

A steel bomb with a volume of 0.53 m^{3} is filled with carbon dioxide with the amount of substance 1 kmol and the pressure of 5.07 MPa. What is the difference between the temperature of the gas calculated by using the model of an ideal gas and the temperature calculated by using the Van der Waals equation?

#### Numerical values

*V*= 0.53 m^{3}the internal volume of the container *n*= 1 kmol = 10^{3}molthe amount of substance of carbon dioxide *p*= 5.07 MPa = 5.07·10^{6}Pathe pressure of carbon dioxide Δ *T*= ?the difference between the temperatures calculated by each of the models

*From The Handbook of Chemistry and Physics:**R*= 8.31 Jmol^{−1}K^{−1}the molar gas constant *a*= 0.365 Jm^{3}mol^{−2}the "van der Waals" constant for carbon dioxide *b*= 4.28·10^{−5}m^{3}mol^{−1}the "van der Waals" constant for carbon dioxide #### Analysis

We will evaluate the thermodynamic temperature from both the ideal gas equation and the Van der Waals equation and then we will calculate their difference.

#### Solution

First we evaluate the thermodynamic temperature

\[pV=nRT_{i},\]*T*_{i}from the ideal gas equationwhere

\[T_{i}=\frac{pV}{nR}.\]*p*is the pressure,*V*is the volume,*n*is the amount of substance of carbon dioxide and*R*is the molar gas constant.We also evaluate the thermodynamic temperature

\[\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)=nRT_{v},\]*T*_{v}from the van der Waals equationwhere

\[T_{v}=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)}{nR}.\]*a*and*b*are the van der Waals constants for carbon dioxide, and thus we obtainFinally, we determine the temperature difference Δ

\[\mathrm{\Delta}T=T_{v}-T_{i}=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)-pV}{nR}.\]*T*:#### Numerical insertion

\[\mathrm{\Delta}T=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)-pV}{nR}\] \[\mathrm{\Delta}T=\frac{\left(5.07\cdot{10^6}+\frac{10^6\cdot{0.365}}{0.53^{2}}\right)(0.53-10^3\cdot{4.28}\cdot{10^{-5}})-5.07\cdot{10^6}\cdot{0.53}}{8.31\cdot{10^3}}\,\mathrm{K}\] \[\mathrm{\Delta}T\dot{=}50\,\mathrm{K}\]#### Answer

The temperature calculated by the van der Waals equation is approximately 50 K higher than the temperature calculated by the ideal gas equation.