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sattari [20]
2 years ago
11

Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a

re the temperature and the specific enthalpy of water at the compressor exit? use steam tables.
Chemistry
1 answer:
Annette [7]2 years ago
6 0
<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:</u></em>

<em><u>s</u></em><em><u>=</u></em><em><u> </u></em><em><u>8</u></em><em><u>.</u></em><em><u>26</u></em><em><u>5</u></em><em><u>2</u></em><em><u> </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kgK</u></em>

<em><u>The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:</u></em>

<em><u>T₂ = T₁+</u></em><em><u> </u></em><em><u>T₂ - </u></em><em><u>T₁</u></em><em><u>/</u></em><em><u> </u></em><em><u>8</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8</u></em><em><u>₁</u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>s</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u>)</u></em>

<em><u>= </u></em><em><u>(</u></em><em><u>400 +</u></em><em><u> </u></em><em><u>500 - 400</u></em><em><u>/</u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u> </u></em><em><u>(8.2652 - 8</u></em><em><u>)</u></em><em><u>)</u></em>

<em><u>= 478.83°C</u></em>

<em><u>The final enthalpy is determined in the same way:</u></em>

<em><u>h₂= h₁</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>h₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>h₁</u></em><em><u>/</u></em><em><u>s</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u> </u></em><em><u>( s - s₁)</u></em>

<em><u>= (</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>+</u></em><em><u> </u></em><em><u>3486.6 </u></em><em><u>-</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>/</u></em><em><u> </u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u>)</u></em><em><u> </u></em><em><u>(8.265</u></em><em><u>)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>3441.91 </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kg</u></em>

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Explanation:

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What happens when balloon filled with CO2 is released in air​
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Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %
Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol

Step 2: Calculate the molar ratio

C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

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