Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.

Answers

Answer 1

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

Related Questions

Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.

Answers

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

Answer: 0.2025

Explanation: I got it correct

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol

Answers

Answer: Option D) 298 g/mol is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol ≈ 298 g/mol

Therefore Option D) 298 g/mol is the correct answer

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