A mixture (air-petrol) undergoes a transformation from the state (1) to the state (2) following three different paths (a, b and c) with: The 1st transformation is isochore then isobaric (path a), the 2nd is isobaric then isochore (path b) and the 3rd is such that PV=cste (path c).

State(1): p1=1bar; v1 = 3l

State(2): p1= 3bar; V1= 1l

1) Represent the three transformations in Clapeyron coordinates.
2) Calculate ΔU between the state (1) and the state (2).

3) Calculate the work in all three cases and deduct the heat exchanged; are they received or rejected

Answer: To represent the three transformations in Clapeyron coordinates, you would need to plot the pressure (p) against the volume (V) for each path. This would involve calculating the intermediate states for each path.

To calculate ΔU between the state (1) and the state (2), you would need to use the internal energy equation, which involves the temperature (T) and the specific heat capacity (c_v) of the mixture.

To calculate the work in each case, you would need to use the work equation, which involves the pressure (p) and the volume (V) changes. Then, you could use the first law of thermodynamics to relate the work to the heat exchanged.

If you're interested in learning more about thermodynamic processes and calculations, I recommend consulting with a specialist in the relevant field or seeking out credible sources of information.

Explanation:

MathPhys MathPhys · Answer 2 · 2024-05-09T10:56:16-04:00

Answer:

1) See graph.

2) ΔU = 0 J

3) Path A: W = 600 J added, Q = -600 J removed

Path B: W = 200 J added, Q = -200 J removed

Path C: W = 330 J added, Q = -330 J removed

Explanation:

1) A Clapeyron coordinate graph is also known as a PV diagram, where pressure is the y-axis and volume is the x-axis. An isochoric process is constant volume, and appears as a vertical line on a PV diagram. An isobaric process is constant pressure, and appears as a horizontal line on a PV diagram. The equation PV = constant describes an isothermal process, or constant temperature, which appears as a curve on a PV diagram.

2) Air can be modeled as a diatomic ideal gas; for diatomic ideal gases, the change in internal energy is:

ΔU = 5/2 nR (T₂ − T₁) = 5/2 (P₂V₂ − P₁V₁)

This change in internal energy is path independent, meaning it is the same regardless of which path is taken between state 1 and state 2.

ΔU = 5/2 [(3 bar) (1 L) − (1 bar) (3 L)]

ΔU = 0 J

3) From the first law of thermodynamics, the sum of the heat added and the work added is equal to the change in internal energy.

Q + W = ΔU

In this case, ΔU = 0, so Q = -W.

Isobaric work is W = -PΔV, isochoric work is W = 0, and isothermal work for an ideal gas is W = -P₁V₁ ln(V₂/V₁) = -P₂V₂ ln(V₂/V₁).

For path A, the work added is:

W = 0 + -(3 bar) (1 L − 3 L) = 6 bar L = 600 J

So the heat added is -600 J (heat removed).

For path B, the work added is:

W = -(1 bar) (1 L − 3L) = 2 bar L = 200 J

So the heat added is -200 J (heat removed).

For path C, the work added is:

W = -(1 bar) (3 L) ln(1 L / 3 L) = 3.30 bar L = 330 J

So the heat added is -330 J (heat removed).