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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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When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest
Aleksandr [31]

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group CH_3.

In the axial position, we have a more steric hindrance because we have two hydrogens near to the CH_3 group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the CH_3 group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>

See figure 1

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6 0
3 years ago
Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry
Ipatiy [6.2K]

<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm

<u>Explanation:</u>

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT

where,

\pi = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of sucrose = 3.40 g

M_{solute} = molar mass of sucrose = 342.3 g/mol

V_{solution} = Volume of solution = 1 L

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 20^oC=[20+273]K=293K

Putting values in above equation, we get:

\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm

Hence, the pressure that must be applied to the apparatus is 0.239 atm

3 0
3 years ago
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<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

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3 years ago
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Why do the Noble gases have the highest ionization energies across a period?
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Answer:

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The answer to this question would be the last option - D) 414
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