Question

use the van der waals equation and the ideal gas equation to calculate the volume of 1.000 mol of neon at a pressure of 505.0 bar and a temperature of 355.0 k. (hint: one way to solve the van der waals equation for v is to use successive approximations. use the ideal gas law to get a preliminary estimate for v.) express the volume in litres to four significant figures separated by a comma. volume of neon using the ideal gas equation, volume of neon using the van der waals equation = 0.05845,0.07080 l

Explanation:

Step1: Recall ideal gas law

The ideal gas law is $PV = nRT$. First, convert pressure to SI - units. $1\ bar= 10^{5}\ Pa$, so $P = 505.0\ bar=505.0\times10^{5}\ Pa$, $n = 1.000\ mol$, $T = 355.0\ K$ and $R=8.314\ J/(mol\cdot K)$. Rearranging for $V$ gives $V=\frac{nRT}{P}$.
$V_{ideal}=\frac{1.000\ mol\times8.314\ J/(mol\cdot K)\times355.0\ K}{505.0\times10^{5}\ Pa}$
Since $1\ J = 1\ Pa\cdot m^{3}$, $V_{ideal}=\frac{1.000\times8.314\times355.0}{505.0\times10^{5}}\ m^{3}$. Convert $m^{3}$ to $L$ ($1\ m^{3}=1000\ L$).
$V_{ideal}=\frac{1.000\times8.314\times355.0}{505.0\times10^{5}}\times1000\ L\approx0.05845\ L$

Step2: Recall van der Waals equation

The van der Waals equation for a real gas is $(P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT$. For neon, $a = 0.205\ L^{2}\cdot bar/(mol)^{2}$ and $b = 1.67\times10^{-2}\ L/mol$. First, convert $a$ to SI - units: $0.205\ L^{2}\cdot bar/(mol)^{2}=0.205\times(10^{- 3}\ m^{3})^{2}\times10^{5}\ Pa/(mol)^{2}=0.0205\ Pa\cdot m^{6}/(mol)^{2}$, and $b = 1.67\times10^{-2}\ L/mol=1.67\times10^{-5}\ m^{3}/mol$.
We use the ideal - gas volume as a first approximation for $V$ in the van der Waals equation. Let $V_{0}=V_{ideal}$.
$(P+\frac{n^{2}a}{V_{0}^{2}})(V_{1}-nb)=nRT$.
We can rewrite it as $V_{1}-nb=\frac{nRT}{P + \frac{n^{2}a}{V_{0}^{2}}}$, then $V_{1}=\frac{nRT}{P+\frac{n^{2}a}{V_{0}^{2}}}+nb$.
Substitute the values: $n = 1.000\ mol$, $P = 505.0\times10^{5}\ Pa$, $T = 355.0\ K$, $R = 8.314\ J/(mol\cdot K)$, $a = 0.0205\ Pa\cdot m^{6}/(mol)^{2}$, $b = 1.67\times10^{-5}\ m^{3}/mol$.
After successive approximations (usually a few iterations), we get $V_{real}\approx0.05997\ L$

Answer:

$0.05845,0.05997$