Carnot Cycle and Adiabatic process

Bob19

I have a question regarding Carnot Cycles and Adiabatic processs.

I have cylinder with n moles of a mono-atomic gas with a specific Volume, Pressure and Temperature. This Cylinder is closed by a piston.

This system undergoes the four step Carnot Cycle A-B, B-C, C-D and D-A.

A-B: Is the adiabatic process where no heat gets in or out Q=0. What I'm required to do is to calculate the change in Pressure, Volume and Temperature. Any hits or idears on how I do that?

Is it something about calculating the work done by the gas? Which during the adiabatic process expands the initial volume. But does this also change the pressure and temperature of the gas too ?

Sincerley and Best Regards,

Bob

Bob19

Thanks,

I will this out, and get back to You if the need more asistance.


/BOB

Andrew Mason

You will have to use the adiabatic condition: [tex]PV^\gamma = constant[/tex]

AM

Bob19

Hello again,

The mono-atomic gas is not specified. Therefore which [tex]\gamma [/tex] -value do I use ? 1.64 or 1.69 ??

/Bob

Yegor

i believed that for mono-atomic gas we assume [tex]\gamma=5/3=1.67 [/tex]

Bob19

I have a second question.

As part of the carnot cycle I have the isothermical process D -> C, where

[tex]P_D = 711.3 \textrm{KPa} , V_D = 15.0 \textrm{L and} \ T_D = 584.43 \textrm{K}[/tex]

Since D and C are on the same path in the carnot cycle then according to theory T_D = T_C.

I'm required to calculate V_C from the data availible to me. Which formula do I use to calculate V_C ?

Sincerely
Bob

Yegor

If A-B is adiabatic then C-D (or D-C) is also adiabatic.
Nevertheless, if both are isothermic, then you should use [tex]PV^\gamma = constant[/tex] for adiabatic and [tex]PV = constant[/tex] for isothermic.

I think that any parameter in point B or A must be given. Isn't it?

Bob19

Since A and B are on the same path in the carnot cycle then

[tex]T_A = T_B[/tex]

I have reached the following point (where I'm stuck in my calculation of [tex]V_C[/tex] and [tex]P_{C}[/tex]

[tex]P_{c} * V_{c} = T_{c} \frac{P_{D} * V_{D}}{T_c}[/tex]

I get that [tex]P_{c} * V_{c} = 10671,5[/tex]

Can I use this result to obtain [tex]V_{C}[/tex] and [tex]P_{D}[/tex] ??

/Bob

Yegor

Or it was another problem???
Yes, you can use, but it isn't enough.
Do you realy have nothing more given?

P.S. How it's possible to quote? What is the right command?

Bob19

Hello again,

Yes its a different part of the problem this time:

First I made a typoo:

Here is the right formula

[tex]P_{c} * V_{c} = T_{c} \frac{P_{D} * V_{D}}{T_D}[/tex]

I'm told that [tex]V_C = 24 \ Liters[/tex] but how thats derrived from the above result I'm a bit unsure of ?

/Bob

Yegor

May be you should use any results from another parts of the problem!?
If you have worked this out you should have some information about other points of the cycle.

Bob19

I have data for D and A the volume in C but not other data.

So what do I do ?

BOB

Yegor

C-D and A-B are isothermic; and D-A and B-C - adiabatic.
Then [tex] V_C P_C= V_D P_D[/tex]
[tex] V_A P_A= V_B P_B[/tex]
[tex]P_D V_D^\gamma=P_A V_A^\gamma[/tex]
[tex]P_B V_B^\gamma=P_C V_C^\gamma[/tex]
Also use [tex]PV^\gamma = TV^{\gamma-1}=P^{1-\gamma}T^\gamma =constant[/tex] for adiabatic. Now you can calculate all data

Bob19

Hi,

Which of these formulas do I then use to calcualte V_c ?

/Bob

Yegor

Hm. I think it's impossible. Imagine the P-V plot. You have two points D and A, as you told. There are no restrictions on where adiabata B-C intersects isothermas.

Bob19

Sorry scratch that:

I have discovered that [tex]V_C = 24[/tex] is avalible in the text.

Sorry I missed that cause I'm suffering from terrible hayfever today.

I'm Sorry :-)

/Bob