Answer :
Answer:
a) [tex]P = 1588.329\,kPa[/tex], greater than experimental value. b) [tex]P = 1657.073\,kPa[/tex], greater than experimental value. c) [tex]P = 1364.841\,kPa[/tex], lesser than experimental value.
Explanation:
a) The mathematical model of the equation of state is:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T[/tex]
The expression required to determine the pressure is:
[tex]P = \frac{m\cdot R_{u}\cdot T}{V\cdot M}[/tex]
[tex]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} )}[/tex]
[tex]P = 1588.329\,kPa[/tex]
Which is greater than experimental value.
b) The mathematical model of the equation of state is:
[tex]\left(P+\frac{a}{\nu^{2}} \right)\cdot (\nu-b)=\frac{R_{u}\cdot T}{M}[/tex], where:
[tex]a = \frac{27\cdot R_{u}^{2}\cdot T_{cr}^{2}}{64\cdot P_{cr}\cdot M^{2}}[/tex] and [tex]b = \frac{R_{u}\cdot T_{cr}}{8\cdot M\cdot P_{cr}}[/tex]
Then:
[tex]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}}[/tex]
[tex]a = 1.383\times 10^{-3}\,\frac{kPa\cdot m^{6}}{kg^{2}}[/tex]
[tex]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)}[/tex]
[tex]b = 1.381\times 10^{-3}\,\frac{m^{3}}{kg}[/tex]
The specific volume is:
[tex]\nu = \frac{3.27\,m^{3}}{100\,kg}[/tex]
[tex]\nu = 32.7\times 10^{-3}\,\frac{m^{3}}{kg}[/tex]
Finally, the pressure is cleared in the equation:
[tex]P=\frac{R_{u}\cdot T}{M\cdot(\nu-b)}-\frac{a}{\nu^{2}}[/tex]
[tex]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}}[/tex]
[tex]P = 1657.073\,kPa[/tex]
Which is greater than experimental value.
c) The mathematical model of the equation of state is:
[tex]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}}[/tex]
Where [tex]A = A_{o}\cdot \left(1-\frac{a}{\bar \nu} \right)[/tex] and [tex]B = B_{o}\cdot (1-\frac{b}{\bar \nu} )[/tex].
The specific molar volume of the nitrogen is:
[tex]\bar \nu=\frac{(3.27\,m^{3})}{\frac{100\,kg}{28.013\,\frac{kg}{kmol}} }[/tex]
[tex]\bar \nu = 0.916\,\frac{m^{3}}{kmol}[/tex]
[tex]a = 0.02617, b = -0.00691[/tex]
[tex]A_{o} = 136.2315, B_{o} = 0.05046[/tex]
[tex]A = 136.2315\cdot \left(1 - \frac{0.02617}{0.916} \right)[/tex]
[tex]A = 132.339[/tex]
[tex]B = 0.05046\cdot \left[ 1 - \frac{(-0.00691)}{0.916} \right][/tex]
[tex]B = 0.05084[/tex]
[tex]c = 4.20\times 10^{4}[/tex]
Finally, the pressure is determined:
[tex]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}}[/tex]
[tex]P = 1364.841\,kPa[/tex]
Which is lesser than experimental value.
If a 23.27 m 3 tank is having a 100 kg of nitrogen at the 175K. The tank pressure will be determined using the equation.
- The ideal gas equation, and the van der Waals . The beats and Bridgeman equation.
- The van der Waals force was named after the Dutch Diderik which refers to the distance dependence integration between the atoms.
- The ideal gas laws refer to the compressibility effects and van der wails force. Takes into account the molecular weight and the molecular volume.
Learn more about the 32.3m tank.
brainly.com/question/13898389.