ChemEng stuff followers

Example - The van der Waals equation for gas temperature, volume and pressure changes

This example was adapted from Levine I., Physical chemistry.

The situation

A sample of 65 mg of $CO_2$ gas at 0.800 bar pressure has its volume 25 dm$^3$ doubled and its absolute temperature tripled. Find the final pressure in Pa.

Consider that the gas behaves according to the van der Waals equation.

The solution

As you can easily observe from the statement above: pressure, temperature and volume change from one state to another. Let us say,

$P,V,T \rightarrow xP, 2V, 3T$        Eq. (01)

From Eq. (01) we may write two van der Waals equations: one for state 1 (rhs of Eq. (01)) and another one for state 2 (lhs of Eq. (01)). These are,

$\left( P+\dfrac{an^2}{V^2} \right) \left( V-nb \right)=nRT$        Eq. (02)
$\left( xP+\dfrac{an^2}{4V^2} \right) \left( 2V-nb \right)=3nRT$        Eq. (03)

where the parameter $x$ has been introduced to account for a possible change in pressure as a multiple of the initial 0.8 bar. Therefore, if we are able to find $x$ all we would need is to multiply it by 0.8 bar.

How to

One way of proceeding is to accept that the mass of the gas is unchanged. In other words the factor $nR$ remains constant for both Eqs (02-03). Next we may isolate this from both equations and make use of this condition as follows

$\dfrac{\left( P+\dfrac{an^2}{V^2} \right) \left( V-nb \right)}{T}=\dfrac{\left( xP+\dfrac{an^2}{4V^2} \right) \left( 2V-nb \right)}{3T}$        Eq. (04)

Further simplification of Eq. (04) leads to,

$\left( P+\dfrac{an^2}{V^2} \right) \left( V-nb \right)=\dfrac{1}{3}\left( xP+\dfrac{an^2}{4V^2} \right) \left( 2V-nb \right)$        Eq. (04)

Isolation of $x$ goes as follows,

$3\left( P+\dfrac{an^2}{V^2} \right) \dfrac{\left( V-nb \right)}{\left( 2V-nb \right)} =\left( xP+\dfrac{an^2}{4V^2} \right) $        Eq. (05)

$x$ is finally found to be,

$x = \dfrac{1}{P}\left[ 3\left( P+\dfrac{an^2}{V^2} \right) \dfrac{\left( V-nb \right)}{\left( 2V-nb \right)} - \dfrac{an^2}{4V^2}\right] $        Eq. (05)

Next, for $CO_2$ gas $M=$ 44.01 g/mole so that $n=$ 1.48 mole. Using SI units $P=$ 80,000 Pa and $V=$ 0.025 m$^3$. Also, the van der Waals equation constants for this gas are: $a=0.36$ m$^6$ Pa/mol$^2$ and $b=4.27\times 10^{-5}$ m$^3$/mol. Therefore, substitution of the known data into Eq. (05) gives,

$x=1.52$

This means that the pressure increases by $\approx 3/2$. The new value of the pressure is about 121,422.83 Pa.

A short discussion

As mentioned early, this example was based on an exercise form the a texbook. That exercise is presented for an unknown ideal gas and the change in pressure is exacly $x=$ 3/2.

As you can see the very same case work out using the van der Waals equation demands more specific data but the result is similar.

No comments:

Post a Comment

Subscribe to: Posts (Atom)

Most popular posts