Question

A 3.27-m3 tank contains 100 kg of nitrogen at 175 K. Determine the pressure in the tank using (a) the ideal-gas equation, (b) the van der Waals equation, and (c) the Beattie- Bridgeman equation. Compare your results with the actual value of 1505 kPa.

Answer 1

Answer:

a) $P = 1588.329\,kPa$, greater than experimental value. b) $P = 1657.073\,kPa$, greater than experimental value.  c) $P = 1364.841\,kPa$, lesser than experimental value.

Explanation:

a) The mathematical model of the equation of state is:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

The expression required to determine the pressure is:

$P = \frac{m\cdot R_{u}\cdot T}{V\cdot M}$

$P = \frac{(100\,kg)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (175\,K)}{(3.27\,m^{3})\cdot (28.013\,\frac{kg}{kmol} )}$

$P = 1588.329\,kPa$

Which is greater than experimental value.

b) The mathematical model of the equation of state is:

$\left(P+\frac{a}{\nu^{2}} \right)\cdot (\nu-b)=\frac{R_{u}\cdot T}{M}$, where:

$a = \frac{27\cdot R_{u}^{2}\cdot T_{cr}^{2}}{64\cdot P_{cr}\cdot M^{2}}$ and $b = \frac{R_{u}\cdot T_{cr}}{8\cdot M\cdot P_{cr}}$

Then:

$a = \frac{27\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )^{2}\cdot (126.2\,K)}{64\cdot(3390\,kPa)\cdot(28.013\,\frac{kg}{kmol} )^{2}}$

$a = 1.383\times 10^{-3}\,\frac{kPa\cdot m^{6}}{kg^{2}}$

$b = \frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (126.2\,K)}{8\cdot (28.013\,\frac{kg}{kmol} )\cdot (3390\,kPa)}$

$b = 1.381\times 10^{-3}\,\frac{m^{3}}{kg}$

The specific volume is:

$\nu = \frac{3.27\,m^{3}}{100\,kg}$

$\nu = 32.7\times 10^{-3}\,\frac{m^{3}}{kg}$

Finally, the pressure is cleared in the equation:

$P=\frac{R_{u}\cdot T}{M\cdot(\nu-b)}-\frac{a}{\nu^{2}}$

$P=\frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (175\,K)}{(28.013\,\frac{kg}{kmol} )\cdot (32.7\times 10^{-3}\,\frac{m^{3}}{kg}-1.381\times 10^{-3}\,\frac{m^{3}}{kg})} -\frac{1.383\cdot 10^{-3}\,\frac{kPa\cdot m^{6}}{kg^{2}}}{(32.7\times 10^{-3}\,\frac{m^{3}}{kg} )^{2}}$

$P = 1657.073\,kPa$

Which is greater than experimental value.

c) The mathematical model of the equation of state is:

$P = \frac{R_{u}\cdot T}{\bar \nu^{2}}\cdot \left( 1 - \frac{c}{\nu\cdot T^{3}} \right)\cdot \left(\bar \nu + B \right) - \frac{A}{\bar \nu^{2}}$

Where $A = A_{o}\cdot \left(1-\frac{a}{\bar \nu} \right)$ and $B = B_{o}\cdot (1-\frac{b}{\bar \nu} )$.

The specific molar volume of the nitrogen is:

$\bar \nu=\frac{(3.27\,m^{3})}{\frac{100\,kg}{28.013\,\frac{kg}{kmol}} }$

$\bar \nu = 0.916\,\frac{m^{3}}{kmol}$

$a = 0.02617, b = -0.00691$

$A_{o} = 136.2315, B_{o} = 0.05046$

$A = 136.2315\cdot \left(1 - \frac{0.02617}{0.916} \right)$

$A = 132.339$

$B = 0.05046\cdot \left[ 1 - \frac{(-0.00691)}{0.916} \right]$

$B = 0.05084$

$c = 4.20\times 10^{4}$

Finally, the pressure is determined:

$P =\frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (175\,K)}{0.916\,\frac{m^{3}}{kmol} }\cdot \left[1 - \frac{4.20\times 10^{4}}{(0.916\,\frac{m^{3}}{kmol} )\cdot (175\,K)^{3}} \right]\cdot (0.916\,\frac{m^{3}}{kmol} + 0.05084)-\frac{132.339}{(0.916\,\frac{m^{3}}{kmol} )^{2}}$

$P = 1364.841\,kPa$

Which is lesser than experimental value.