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Hunter-Best [27]
3 years ago
7

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) T

he 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.
Engineering
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

2.279 bar

Explanation:

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

Also, you should know that 1 l*bar = 0.1 kJ

Lastly, the definition of work for <em>any</em> thermodynamics process:

W=-\int P \, dV                                                           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

-\int\limits^{V_2} _{V_1} {P} \, dV

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:

T = \frac{PV}{nR}

But there is no T in this integral!

W=-\int P \, dV

This is where we will do a little calculus trick.

In differential form:

dT=d\frac{PV}{R}

Note that P and R is constant So...

dT=\frac{P}{R}dV

Now we can substitute dV in the integral with dT

-\int\limits^{T_2} _{T_1} {\frac{PR}{P}} \, dT

And we get

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

W = -3.741\ kJ

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

P=\frac{RT}{V}

Which we will substitute into our ever reliable eq1

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

Solving this integral, we get

W= -RT(ln(V_2)-ln(V_1))=-RTln(\frac{V_2}{V_1})                            eq2

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

V=\frac{RT}{P}

And substitute it in eq2, so:

W= -RTln(\frac{P_1}{P_2}=-RT(ln(P_1)-ln(P_2))

In kJ unit

W=-0.1*800*0.08314*(ln(4)-ln(P_2))

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

W=-3.741 =-0.1*800*0.08314(ln(4)-ln(P_2))

Solve this equation for P_{2} and we get

P_{2}=2.279\ bar

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