Calculating Work in Isothermal Processes of Carnot Cycle

[Abigale]

Hallo,

I read about thr Carnot-Cycle and got a problem.

A->B: isothermal
B->C: adiabatic
C->D: isothermal
D->A: adiabatic


I want to understand how to calculate the work done in the isothermal processes.
I have read:

$W_{AB}= T_{H}Nk\ln{(\frac{V_{A}}{V_{\text{B}}})}\\ W_{CD}= -T_{n}Nk\ln{(\frac{V_{A}}{V_{\text{B}}})}$

But if i callculate i get:
$W_{CD}= -T_{n}Nk\ln{(\frac{V_{D}}{V_{C}})}$

So the question is, why is:
$(\frac{V_{A}}{V_{\text{B}}}) = (\frac{V_{D}}{V_{C}})$ ?

Thx
Abbigale

 

[Simon Bridge]

So the question is, why is:
$(\frac{V_{A}}{V_{\text{B}}}) = (\frac{V_{D}}{V_{C}})$

... because that's how it comes out when you do the rest of the math.
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm

 

[Andrew Mason]

Hallo,

I read about thr Carnot-Cycle and got a problem.

A->B: isothermal
B->C: adiabatic
C->D: isothermal
D->A: adiabatic


I want to understand how to calculate the work done in the isothermal processes.
I have read:

$W_{AB}= T_{H}Nk\ln{(\frac{V_{A}}{V_{\text{B}}})}\\ W_{CD}= -T_{n}Nk\ln{(\frac{V_{A}}{V_{\text{B}}})}$

But if i callculate i get:
$W_{CD}= -T_{n}Nk\ln{(\frac{V_{D}}{V_{C}})}$

So the question is, why is:
$(\frac{V_{A}}{V_{\text{B}}}) = (\frac{V_{D}}{V_{C}})$ ?

Abigale,

From the first law we know that ΔU + W = Q where W is the work done BY the gas. But since ΔU = 0 in one complete cycle, W = Q = heat flow in from hot register - heat flow out to cold register = |Qh| - |Qc|. Heat flow occurs only from A-B and C-D, so the difference in these heat flows is the work done for the entire cycle. That means the work done in the adiabatic parts nets out to 0.

The efficiency is:

$$\eta = 1 - \frac{|Q_c|}{|Q_h|} = 1 - \frac{T_cNk\ln\left(\frac{V_D}{V_C}\right)}{T_hNk\ln\left(\frac{V_A}{V_B}\right)}$$

But we also know that for a Carnot cycle:

ΔS = Qh/Th + Qc/Th = 0 so |Qc/Qh| = Tc/Th

This necessarily means that $(\frac{V_{A}}{V_{\text{B}}}) = (\frac{V_{D}}{V_{C}})$

AM

 

[Abigale]

Hey,
but is it possible to schow the Relation, just by looking at the pv-diagram?

 

[Andrew Mason]

Hey,
but is it possible to show the Relation, just by looking at the pv-diagram?

Yes, if the PV graph showed the scale.

AM

 

[Philip Wood]

The equations you give for isothermal work are for an ideal gas.

Since we're dealing with an ideal gas we can use the adiabatic equations for an ideal gas, for BC and DA.

So $p_B V_B^{\gamma} = p_C V_C^{\gamma}$

and $p_A V_A^{\gamma} = p_D V_D^{\gamma}$

dividing: $\frac{p_B}{p_A} \frac{V_B^{\gamma}}{V_A^{\gamma}} = \frac{p_C}{p_D} \frac{V_C^{\gamma}} {V_D^{\gamma}}$

But for the isothermals: $\frac{p_B}{p_A} = \frac{V_A}{V_B}$ and $\frac{p_C}{p_D} = \frac{V_D}{V_C}.$

We can now eliminate the pressure ratios from the previous equation, giving...

$\frac{V_B^{\gamma - 1}}{V_A^{\gamma - 1}} = \frac{V_C^{\gamma - 1}} {V_D^{\gamma - 1}}$

So $\frac{V_B}{V_A} = \frac{V_C} {V_D}$

 

[Andrew Mason]

The equations you give for isothermal work are for an ideal gas.

Since we're dealing with an ideal gas we can use the adiabatic equations for an ideal gas, for BC and DA.

So $p_B V_B^{\gamma} = p_C V_C^{\gamma}$

and $p_A V_A^{\gamma} = p_D V_D^{\gamma}$

dividing: $\frac{p_B}{p_A} \frac{V_B^{\gamma}}{V_A^{\gamma}} = \frac{p_C}{p_D} \frac{V_C^{\gamma}} {V_D^{\gamma}}$

But for the isothermals: $\frac{p_B}{p_A} = \frac{V_A}{V_B}$ and $\frac{p_C}{p_D} = \frac{V_D}{V_C}.$

We can now eliminate the pressure ratios from the previous equation, giving...

$\frac{V_B^{\gamma - 1}}{V_A^{\gamma - 1}} = \frac{V_C^{\gamma - 1}} {V_D^{\gamma - 1}}$

So $\frac{V_B}{V_A} = \frac{V_C} {V_D}$

It might be easier to use the adiabatic condition in terms of temperature and volume:

$TV^{\gamma - 1} = K$ from which it is apparent that adiabatic paths between the same two temperatures result in the same proportional changes in volume.

AM

 

[Philip Wood]

Much neater – if you can remember TV^γ-1 = K. [Though, I suppose TV^anything = K would suffice!]

We've got to be careful of the logic here. OP's argument leads to $\left| \frac{Q_{AB}}{Q_{CD}} \right| = \frac{T_{AB}}{T_{CD}}$. The temperatures which appear in this equation are (in my interpretation of the above argument) ideal gas scale temperatures, effectively defined by pV = nRT. But the Second Law leads to a definition of thermodynamic temperature based on Carnot cycles for any working substance (not just an ideal gas) that is essentially $\frac{T_{AB}}{T_{CD}} = \left| \frac{Q_{AB}}{Q_{CD}} \right|$. Hence the OP's argument shows the equivalence of the thermodynamic scale and the ideal gas scale.