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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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An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar
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Explanation:

First, we will calculate fuel consumption is as follows.

         372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s

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Now, we will calculate the power as follows.

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5 0
3 years ago
What is the value for AG at 5000 K if AH = -220 kJ/mol and S= -0.05 kJ/(mol-K)?
Serhud [2]

Answer:

C. 30 kJ

Explanation:

Hello there!

In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

\Delta G= \Delta H-T\Delta S

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

\Delta G=-220kJ/mol-5000K*-0.05kJ/mol*K\\\\\Delta G=30kJ/mol

Therefore, the answer is C. 30 kJ .

Best regards!

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2 years ago
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Answer:

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Hey There!

Here is your answer:

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