Physics High School

Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigid tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C.
initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?

Answers

Answer 1

Answer:

Explanation:

For entropy change the formula is

ΔS = ΔQ / T

ΔQ = Δ H

ΔS = Δ H / T

Given

Δ H = + 162.8 kJ

We can take equilibrium temperature as average temperature of the whole process

So, T = 273 + 87 = 360 K

ΔS = Δ H / T

=  162.8 kJ  / 360

= +  0.508 kJ / K .

Answer 2

When the magnitude of the heat transfer for the process is 162.8 kJ, Then the entropy change for the mixing process, in kJ/K is = + 0.508 kJ / K

What is Entropy change?

For The entropy change, the formula is

Then ΔS = ΔQ / T

After that ΔQ = Δ H

Then ΔS = Δ H / T

Given as per question are:

Then Δ H = + 162.8 kJ

Now We can take equilibrium temperature as average temperature of the whole process are:

So, T is = 273 + 87 = 360 K

Then ΔS = Δ H / T

After that = 162.8 kJ / 360

Therefore, = + 0.508 kJ / K.

Find more information about Entropy change here:

https://brainly.com/question/17241209

Related Questions

A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass

Answers

Answer:

λ₀ = 2 d n

Explanation:

A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.

Let's analyze the general conditions of this interference,

* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º

* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.

          λ = λ₀ / n

where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains

         2d = (m + ½) λ

the expression for destructive interference remains

         2d = m λ

          2d = m λ₀ / n

When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.

Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.

Find us the wavelength that this constructive interference creates

           2d n = m λ₀

           λ₀ = 2 d n / m

To find the minimum wavelength, suppose we observe the first interference pattern m = 1

         λ₀ = 2 d n

where d is the thickness of the film and n the index of refraction of the same

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