A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.

Answers

Answer 1

Answer:

below

Explanation:

A Heat Engine With 0.100 Mol Of A Monatomic Ideal Gas Initially Fills A 3000 Cm3 Cylinder At 800 K. The
Answer 2

Part A) This engine works  per cycle is 254.9 J.

Part B) The thermal efficiency is  23.42%

What is the thermal efficiency?

The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.

For isothermal expansion, work done is

W₁ =nRT₁ x ln(V₂/V₁)

W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)

W₁ = 339.8 J =Q₁

For isochoric cooling ,

W₂ =0

Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)

Q₂ = -748.3 J

For isothermal compression,

W₃ =nRT₂ ln (V₄/V₃)

W₃  = 0.1 x 8.314 x 200 x ln(3000/5000)

W₃ = -84.9J

For isochoric heating

W₄ =0

Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)

Q₄  = -748.3 J

Total work done in all the process W = W₁ +W₂ +W₃ +W₄

W =254.9 J

Thus, the work done is 254.9 J

Thermal efficiency = Work done/Heat taken

η = W/ Q₁ +Q₄

η = [254.9 / 339.8 +748.3 ] x 100 %

η = 0.2342 x 100 %

η = 23.42%

Thus, the thermal efficiency is 23.42%

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Related Questions

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?


16s

300s

15s

23s​

Answers

300 because the mass and weight

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

State Work - Energy Theorem.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

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For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2

Answers

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light

Answers

Answer:

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Explanation:

The energy of a photon is calculated using the following equation;

E = hf

where;

h is Planck's constant = 6.63 x 10⁻³⁴ Js

f is frequency of the photon

[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]

[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

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