1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Fynjy0 [20]
3 years ago
6

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

L21 5 1.1. Estimate the value of the parameters at 508C
Chemistry
1 answer:
Gala2k [10]3 years ago
8 0

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C   and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound              Liquid molar volume    (cm³/mol)

Ethanol (1)                    58.68

1 - propanol (2)            75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = \frac{V_2}{V_1} \  exp \ (\frac{-a_{12}/R}{T} )          ------------ equation (1)

where:

a_{12}/R = Wilson parameter = ???

V_2 = liquid molar volume of component 2 = 75.14 cm³/mol

V_1 = liquid molar volume of component 1  = 58.68 cm³/mol

T = temperature  = 25° C  = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

In (0.547) =  \ (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R=   0.60 * 298.15 \ K

-a_{12}/R=   - 178.89 \ K

a_{12}/R=    178.89 \ K

To calculate the temperature dependent parameters of the Wilson equation  ∧₂₁

∧₂₁ = \frac{V_1}{V_2} \  exp \ (\frac{-a_{12}/R}{T} )          ---------- equation (2)

1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

1.1 = 0.7809 \ exp \  (\frac{-a_{12}/R}{298.15 \ K} )

\frac{1.1}{0.7809}=    exp \  (\frac{-a_{12}/R}{298.15 \ K} )

1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R =     0.3426 * 298.15 \ K

-a_{12}/R =102.15 \ K

a_{12}/R = -102.15 \ K

From equation (1) ; let replace  178.98 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )

∧₂₁ =  0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

You might be interested in
How to find the cube root of 1331 ​
Triss [41]
Answer :=11
Thank me later
8 0
3 years ago
The colour imparted to a flame by calcium ion?
igor_vitrenko [27]
I not sure, but i know that is soft yellow no white not yellow.
7 0
3 years ago
Which substance can lead to addiction, or drug dependence, if abused?
worty [1.4K]

Answer:

D

Explanation:

All of those can produce happy chemicals which cause your brain to become dependent. The side effects can be life-threatening but if it makes your brain irrationally happy then it doesn't matter

7 0
3 years ago
4.
Evgen [1.6K]

Answer:

λ = 2.8 m

Explanation:

Given data:

Frequency of radio wave = 106.7 ×10⁶ Hz

Wavelength of radio wave = ?

Solution:

Formula:

Speed of wave = frequency  × wavelength

speed of wave = 3×10⁸ m/s

by putting values,

3×10⁸ m/s = 106.7 ×10⁶ Hz × λ

Hz = s⁻¹

λ = 3×10⁸ m/s / 106.7 ×10⁶ Hz

λ =  3×10⁸ m/s / 106.7 ×10⁶ s⁻¹

λ =  0.028×10² m

λ = 2.8 m

7 0
2 years ago
A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?
ss7ja [257]
O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol

4 0
3 years ago
Read 2 more answers
Other questions:
  • Identify the nuclide produced when plutonium-239 decays by alpha emission: 239 94pu→42he + ? express your answer as an isotope u
    11·1 answer
  • Using the equation, 4Fe + 3O2 Imported Asset 2Fe2O3, if 8 moles of iron and oxygen from the air were available, how many moles o
    5·2 answers
  • An atom of gold is made up of fewer protons than an atom of oxygen? True or False?
    10·2 answers
  • A 130.3 g piece of copper (specific heat 0.38 J/g・°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The w
    7·1 answer
  • Which statement best describes the compressibility of a gas?
    10·1 answer
  • if the force between two charged particles increases,but the distance between them does not change, then:
    12·1 answer
  • A chemical equation is balanced when the number of each
    10·1 answer
  • Two disadvantages to deforestation
    6·2 answers
  • How many milliliters are in 8.12 moles of N2
    9·1 answer
  • What is the percent of oxygen in the compound Mnz(S2O3)3​
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!