Respuesta :
Answer:
[tex]Q=25\ kJ[/tex]
[tex]\Delta s= 0.2885 J.K^{-1} [/tex]
Explanation:
Given:
- Initial pressure of nitrogen, [tex]P_1=6\times 10^5\ Pa[/tex]
- initial temperature, [tex]T_1=247+273=520\ K[/tex]
- polytropic index, [tex]n=1.2[/tex]
- initial volume, [tex]V_1=0.1\ m^3[/tex]
- work done in the process, [tex]W=50000\ J[/tex]
For heat interaction during the polytropic process we have:
[tex]Q=W[\frac{\gamma -n}{\gamma-1} ][/tex]
[tex]Q=50\times[\frac{1.4-1.2}{1.4-1} ][/tex]
[tex]Q=25\ kJ[/tex]
For ideal gas we have the Gas Law:
[tex]P_1.V_1=m.R.T_1[/tex]
[tex]6\times 10^5\times 0.1=m.R\times 520[/tex]
[tex]m.R=115.385\ J.K^{-1}[/tex]
For work we have the relation:
[tex]W=m.R.\frac{(T_1-T_2)}{(n-1)}[/tex]
putting respective values
[tex]50000=115.385\times \frac{(520-T_2)}{(1.2-1)}[/tex]
[tex]T_2=433.33\ K[/tex]
We know entropy change:
[tex]\Delta s=\frac{dQ}{dT}[/tex]
[tex]\Delta s=\frac{25}{520-433.33}[/tex]
[tex]\Delta s= 0.2885 J.K^{-1} [/tex]
The heat transfer will be equal to [tex]25 KJ[/tex], while the entropy change will be [tex]0.2885J.K^-^1[/tex].
How to get to this result?
- First, we need to know how heat interacts in the polytropic processor. For this, we must use the equation:
[tex]Q=W[\frac{y-n}{y-1}]\\Q= 50*\frac{1.4-1,2}{1,4-1} \\Q= 25KJ[/tex]
- It is also necessary to know the interaction of heat in an ideal gas. For this we will use the gas law equation, as follows:
[tex]P_1V1=n.r.t_1\\(6*10^5)**0.1=n*r*520\\n*r= 115.385 J.K^-^1[/tex]
- Next, we must consider the behavior of heat to generate work. This will be done with the equation:
[tex]W= n*r*\frac{t_1-t_2}{1.2-1} \\50000=115.385*\frac{520-t_2}{1.2-1} \\t_2= 433.33K[/tex]
- Finally, we can calculate the entropy change with the equation:
[tex]\Delta s= \frac{dQ}{dT} \\\Delta s= \frac{25}{520-433.33} \\\Delta s= 0.2885J. K^-^1[/tex]
More info on entropy at the link:
https://brainly.com/question/15025401
