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sleet_krkn [62]
3 years ago
6

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperatu

re of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you fill in the following table. P V T A 1 400 kPa 10.0 L 720 K B C 24 L D 15 L (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A ? B, B ? C, C ? D, and D ? A. Process Q (kJ) W (kJ) ?Eint (kJ) A ? B B ? C C ? D D ? A (c) Calculate the efficiency Wnet / |Qh|. % (d) Show that the efficiency is equal to 1 ? TC / TA, the Carnot efficiency. (Do this on paper. Your instructor may ask you to turn in this work.)

Physics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

Find attached the solution

Explanation:

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550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. Wh
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Answer:

69.7 cm

Explanation:

What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m

Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ

So, d/mλ = x/D for a second order bright fringe, m = 2.

So, d/2λ = x/D

x = dD/2λ

So, x =

For a dark fringe, we have

d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

d/(2 + 1/2)λ = x'/D

d/(5/2)λ = x'/D

2d/5λ = x'/D

x' = 2dD/5λ

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ

x" = dD/10λ

Substituting the values of the variables into the equation, we have

x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)

x" = 1.15/165 × 10² m

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

6 0
3 years ago
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