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Serga [27]
3 years ago
13

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to the g

as during the isothermal expansion phase of each cycle is indicated.
Chemistry
1 answer:
Alla [95]3 years ago
7 0

Answer:

<u>Part -A</u>

The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

Change in entropy over a complete cycle is 1 =2 = 3= 4  = 5 = 6

Explanation:

<u>Part -A</u>

The change in entropy for the reversible process that transfers the heat energy 'Q' at temperature 'T' is

\Delta S = \frac{Q}{T}

1)

\Delta S_{1} = \frac{Q}{T}

= \frac{1000J}{600K}= 1.67J/K

2)

\Delta S_{2} = \frac{Q}{T}

= \frac{1000J}{500K}= 2J/K

3)

\Delta S_{3} = \frac{Q}{T}

= \frac{1500J}{600K}= 2.5J/K

4)

\Delta S_{4} = \frac{Q}{T}

= \frac{1000J}{500K}= 2J/K

5)

\Delta S_{5} = \frac{Q}{T}

= \frac{500J}{500K}= 1J/K

6)

\Delta S_{6} = \frac{Q}{T}

= \frac{500J}{600K}= 0.83J/K

Therefore, The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

\Delta S = \frac{Q}{T}

Q_{system}=work\,\,done

Initial and final states of complete cycle are equal. after complete the one cycle the system returns to the original state. So, the change in the entropy will be the zero

Therefore, Change in the overall cycle is 1 =2 = 3= 4  =5=6

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One mole of acetyl chloride was mixed with one mole of dimethylamine. After the reaction is complete, what species can be found
fredd [130]

Answer:

N,N-dimethylacetamide is formed.

Explanation:

  • It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
  • In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
  • In the second step, removal of Cl atoms occurs.
  • In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
  • Full reaction mechanism has been shown below.

7 0
3 years ago
A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of
VikaD [51]

Answer:

Mole Fraction (H₂O)  =  0.6303

Mole Fraction (C₂H₅OH)  =  0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O            1 mole
---------------------  x  ------------------  =  11.10 moles H₂O
                                 18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH              1 mole
----------------------------  x  --------------------  =  6.512 moles C₂H₅OH
                                         46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

                                            moles solute
Mole Fraction  =  ------------------------------------------------
                               moles solute + moles solvent

                                                  11.10 moles H₂O
Mole Fraction  =  -------------------------------------------------------------
                               11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O)  =  0.6303

                                             6.512 moles C₂H₅OH
Mole Fraction  =  -------------------------------------------------------------
                               11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH)  =  0.3697

7 0
1 year ago
The image shows a piston system where gas is compressed. If the uncompressed system is at a standard pressure of 1 atm, what is
MrMuchimi

The answer is 2 atm. I guessed on the problem and got it right. I don’t know the work though.

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Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
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D. Sample C would be best
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ASAP Would you expect the odor of an open bottle of perfume to spread through the air faster when the room temperature is 85F or
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Answer: The answer would be 85F  

Explanation: The odour molecules become airborne much more quickly in a warmer environment than a colder one. So the bottle of perfume to spread faster through the air faster when the room temperature is 85F

8 0
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