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A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bar is cooled at constant volume to T = 350 K. (2) The air is then heated at constant pressure until its temperature reaches 800 K. If this two-step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to some final pressure P, what is the value of P that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with CP = (7/2)R and CV = (5/2)R.

Respuesta :

ElennaF

Answer:

2.279 bar

Explanation:

I will be using R = [tex] 8.314*10^{-2} \frac{l*bar}{K*mol}[/tex] here.

Also, you should know that 1 [tex]l*bar[/tex] = 0.1 kJ

Lastly, the definition of work for any thermodynamics process:

[tex]W=-\int P \, dV[/tex]                                                           eq1        

So, getting started. We will first find the work for the first process

  • The first step, we have a constant volume process, therefore, there is no work done (W = 0) You can see it from this integral

[tex]-\int\limits^{V_2} _{V_1} {P} \, dV[/tex]

There is no volume change, so no work can be done

  • In the second step We now have a constant pressure process. Don't worry, our eq1 still works here. Take in mind now that P is now a function of T. And since we are assuming that air is an ideal gas, we can use Ideal gas law PV = nRT here. So:

[tex]T = \frac{PV}{nR}[/tex]

But there is no T in this integral!

[tex]W=-\int P \, dV[/tex]

This is where we will do a little calculus trick.

In differential form:

[tex]dT=d\frac{PV}{R}[/tex]

Note that P and R is constant So...

[tex]dT=\frac{P}{R}dV[/tex]

Now we can substitute dV in the integral with dT

[tex]-\int\limits^{T_2} _{T_1} {\frac{PR}{P}} \, dT[/tex]

And we get

[tex]W = -\int\limits^{T_2} _{T_1} {\frac{PR}{P}} \, dT[/tex]

[tex]W = -3.741\ kJ[/tex]

So, in total, the work done by this process is 3.741 kJ. Make sure you do not confuse the sign of your answer. It is always good to state prior to your calculation which sign, + or - , will be assigned to work done by your system or work done to your system.

Now for the second process

  • We have an isothermal process, which means that T is constant during the process. So, similar to part 2 of the first process, by using ideal gas law:

[tex]P=\frac{RT}{V}[/tex]

Which we will substitute into our ever reliable eq1

[tex]W=-\int\limits^{V_2} _{V_1} {\frac{RT}{V}} \, dV[/tex]

Solving this integral, we get

[tex]W= -RT(ln(V_2)-ln(V_1))=-RTln(\frac{V_2}{V_1})[/tex]                            eq2

Now, since we don't know V but we know P, We can simply use Ideal gas law(again)

[tex]V=\frac{RT}{P}[/tex]

And substitute it in eq2, so:

[tex]W= -RTln(\frac{P_1}{P_2}=-RT(ln(P_1)-ln(P_2))[/tex]

In kJ unit

[tex]W=-0.1*800*0.08314*(ln(4)-ln(P_2))[/tex]

  • Finally, we find [tex]P_2[/tex] of process 2 that would make that work done by both process equal. So, we equate the work done by both process

[tex]W=-3.741 =-0.1*800*0.08314(ln(4)-ln(P_2))[/tex]

Solve this equation for [tex]P_{2}[/tex] and we get

[tex]P_{2}=2.279\ bar[/tex]

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