Question

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?

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.

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