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Ganezh [65]
3 years ago
15

A bottle of 2 moles of an ideal diatomic gas experiences a temperature increase of 50.0K at constant volume. (a) Find the increa

se in internal energy if only translational and rotational motions are possible. (b) Find the increase in internal energy if translational, rotational, and vibrational motions are possible. (c) Which of the following kind of molecule is more difficult to increase temperature with the same amount of heat: (i) molecule that has many vibrational and rotational modes or (ii) molecule with only one of such modes?
Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

Explanation:

a) For diatomic gas: Translational motion = 3 and rotational motion = 2

∴ Total (internal energy) = 3 + 2 = 5

b) Translational + Rotational + Vibrational = 3 + 2 + 1 = 6

c) Linear molecule  

i) Non linear molecule

ii)  Monatomic molecule

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(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3

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