Respuesta :
Answer:
[tex] W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ[/tex]
Explanation:
For this case we know the following info :
[tex] V_i = 0.3 m^3 [/tex] represent the initial volume
[tex] V_f = 0.1 m^3[/tex] represent the final volume
We know that the pressure and volume are related with the following expression:
[tex] P = aV^{-2}= \frac{a}{V^2}[/tex]
Where a is a constant given [tex]a = 6.5 Kpa m^6[/tex]
And we need to calculate the work associated to this process.
We have a compression, and by definition the work is defined with the following general expression:
[tex] W = \int_{V_i}^{V_f} P dV[/tex]
If we replace the expression for P we got:
[tex] W = \int_{V_i}^{V_f} a V^{-2} dV[/tex]
If we integrate we got:
[tex] W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}[/tex]
Using the fundamental theorem of calculus we have:
[tex] W = -a (\frac{1}{V_f} -\frac{1}{V_i})[/tex]
And replacing the values we got:
[tex] W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ[/tex]
The work done on carbon dioxide during this process is; -43.33 kJ
What is work done?
For the Polytropic process, we are given the relationship between pressure and volume as; P = aV⁻² = a/V²
To get the work done, we have to integrate the pressure to get;
W = ∫P = ∫a/V² = -a/V
Work done from 0.3 m³ to 0.1 m³, we have;
W = -6.5(¹/₀.₁ - ¹/₀.₃)
W = -43.33 kPam³ = -43.33 kJ
Read more about finding workdone at; https://brainly.com/question/18762601
