Respuesta :
Answer:
Part a)
[tex]V_2 = 0.033 m^3[/tex]
Part b)
[tex]W = -20 kJ[/tex]
Explanation:
Part a)
As we know that the gas equation is given as
[tex]PV^2 = constant[/tex]
so here we know that
[tex]P_1 = 1 bar[/tex]
[tex]P_2 = 9 bar[/tex]
[tex]V_1 = 0.1 m^3[/tex]
Now in order to find the final volume we have
[tex]V_2^2 = \frac{P_1V_1^2}{P_2}[/tex]
[tex]V_2^2 = \frac{1 \times 0.1^2}{9}[/tex]
[tex]V_2 = 0.033 m^3[/tex]
Part b)
Work done in this process is given as
[tex]W = \int P dV[/tex]
[tex]W = \int \frac{k}{V^2} dV[/tex]
[tex]W = \int_{0.1}^{0.033} \frac{k}{V^2} dV[/tex]
[tex]W = -\frac{k}{V_2} + \frac{k}{V_1}[/tex]
[tex]W = - \frac{P_2V_2^2}{V_2} + \frac{P_1V_1^2}{V_1}[/tex]
[tex]W = P_1V_1 - P_2V_2[/tex]
[tex]W = (1\times 10^5)(0.1) - (9 \times 10^5)(0.033)[/tex]
[tex]W = -20 kJ[/tex]
The work is done on the system which is been denoted by the negative symbol while the amount of work done is equal to 19,700 J.
What is the final volume, in m³?
As it is given to us, PV² = constant is the relationship of the piston-cylinder assembly,
Given to us,
P₁ = 1 bar
V₁ = 0.1 m³
P₂ = 9 bar
therefore, substituting the values in the relationship given to us
[tex]P_1V_1^2 = P_2V_2^2[/tex]
[tex]1\times (0.1)^2=9\times(V_2)^2[/tex]
V₂ = 0.0334 m³
Hence, the final volume in the piston-cylinder assembly is 0.0334 m³.
What is the work done by the piston-cylinder assembly?
We know that for closed systems, work done is given as,
[tex]W = \int Pdv[/tex]
therefore,
[tex]W = \int^{V_2}_{V_1} \dfrac{k}{V^2}dv\\\\W = [-\dfrac{k}{V}]^{V_2}_{V_1}\\\\W = [-\dfrac{PV^2}{V}]^{V_2}_{V_1}\\\\W = [-{PV}]^{V_2}_{V_1}\\\\W = -P_2V_2+P_1V_1\\W = -(9 \times 10^5 \times 0.0334)+(1\times 10^5 \times 0.1)\\W = -19700\rm\ J[/tex]
Hence, the work is done on the system which is been denoted by the negative symbol while the amount of work done is equal to 19,700 J.
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