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tatiyna
2 years ago
9

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

) is fed to a condenser at 75 deg C and 3.0 atm absolute. The product gas stream and condensed hexane stream emerge at 20 deg C and 3.0 atm absolute. Calculate the amount of heat flow in kJ required per gmol of hexane condensed. Your answer will be negative.
Chemistry
1 answer:
kondor19780726 [428]2 years ago
6 0

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
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The question is incomplete, the complete question is;

The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.

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-1.67 JK-1

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Since Heat of vaporization of acetic acid = 41.0 kJ/mol

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Heat of condensation of acetic acid = -41.0 kJ/mol

Mass of acetic acid = 954 g

Temperature of condensation = 118.1 °C or 391.1 K

Number of moles of acetic acid = 954 g/60g/mol = 15.9 moles

Heat evolved during condensation = 15.9 moles *  -41.0 kJ/mol = -651.9 KJ

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6.02 × 10²³ atoms.

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The number 6.02 × 10²³ is called Avogadro number. it is the number of atoms, ions or molecules in one gram atoms of element, one gram ions of substance or one gram molecules of a compound.

For one mole of magnesium.

24.305 g of Mg = 1 mole of magnesium =  6.02 × 10²³ atoms of magnesium.

Other example includes:

18 g of water = one mole of water =  6.02 × 10²³ molecules of water.

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What is the partition coefficient?

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Learn more about octanol here:

brainly.com/question/7768749

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