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morpeh [17]
3 years ago
7

Air is compressed in a piston–cylinder device from 90 kPa and 20°C to 650 kPa in a reversible isothermal process. Determine the

entropy change of air. The gas constant of air is R = 0.287 kJ/kg·K. (You must provide an answer before moving on to the next part.)
Chemistry
1 answer:
Sophie [7]3 years ago
4 0

Answer:

-0.5674 \frac{kJ}{kgK}

Explanation:

The entropy change of an ideal gas can be calculated by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

This is an isothermal process, so the entropy change will be calculated as:

s_{2}-s_{1}=-Rln(\frac{P_{2}}{P_{1}})

s_{2}-s_{1}=-0.287*ln(\frac{650}{90})=-0.5674[tex]\frac{kJ}{kgK}[/tex]

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Answer:

answer is b

Explanation:

All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor

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Give three examples of energy being changed to other forms.
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6 0
3 years ago
How many nickel atoms are in a nickel coin with a mass of 3.95g
lyudmila [28]

Answer:

4.05 × 10²² atoms

Explanation:

Step 1: Given data

Mass of nickel: 3.95 g

Step 2: Calculate the moles corresponding to 3.95 g of nickel

The molar mass of nickel is 58.69 g/mol.

3.95 g × (1 mol/58.69 g) = 0.0673 mol

Step 3: Calculate the atoms in 0.0673 moles of nickel

We will use Avogadro's number: there are 6.02 × 10²³ atoms of nickel in 1 mole of atoms of nickel.

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5 0
3 years ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
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