How is this heat engine formula derived?

[hahaha158]

Homework Statement

there is a question solution i am looking at, and it goes from TH(Vb)^(y-1) = TH(Vc)^(y-1) and it appears to be rewritten as Vc=Va(TH/TC)^(1/(y-1))

Can anyone please explain this?

thanks

 

[Simon Bridge]

You mean this:$$T_H V_b^{\gamma -1}=T_H V_c^{\gamma -1}$$ gets you $$V_c=V_a\left ( \frac{T_H}{T_C} \right )^{\frac{1}{\gamma -1}}$$... looks like a heat-engine process and the secret is how state a is related to states b and c. Presumably some relationship with the temperature of the cold reservoir. The first equation is not all they used ... you have to reference the process being described to understand the derivations.

I suspect an earlier step gave you: $$T_C V_b^{\gamma -1}=T_H V_a^{\gamma -1}$$ ... so they just used the first equation to eliminate $V_b$ from that last equation and solved for $V_c$.

 

[hahaha158]

You mean this:$$T_H V_b^{\gamma -1}=T_H V_c^{\gamma -1}$$ gets you $$V_c=V_a\left ( \frac{T_H}{T_C} \right )^{\frac{1}{\gamma -1}}$$... looks like a heat-engine process and the secret is how state a is related to states b and c. Presumably some relationship with the temperature of the cold reservoir. The first equation is not all they used ... you have to reference the process being described to understand the derivations.

I suspect an earlier step gave you: $$T_C V_b^{\gamma -1}=T_H V_a^{\gamma -1}$$ ... so they just used the first equation to eliminate $V_b$ from that last equation and solved for $V_c$.

I get $$T_H V_b^{\gamma -1}=T_C V_c^{\gamma -1}$$ gets you $$V_c=V_a\left ( \frac{T_H}{T_C} \right )^{\frac{1}{\gamma -1}}$$

I do not think there is a previous step that looks like the one you listed.

Edit: The original question statement is

Suppose 0.200 mol of an ideal diatmoic gas (y=1.4) undergoes a carnot cycle between 227 and 27 degrees celsius, starting at pressure A = 10x10^5 Pa at point a in the pV diagram (don't know how to post it). The volume doubles during isothermal expansion step a->b.

Is there anytihn you can take from this that might imply something?

 

[Simon Bridge]

I get $$T_H V_b^{\gamma -1}=T_C V_c^{\gamma -1}$$ gets you $$V_c=V_a\left ( \frac{T_H}{T_C} \right )^{\frac{1}{\gamma -1}}$$

Oh OK ... in that case the previous step would have looked like: $$T_H V_b^{\gamma -1}=T_H V_a^{\gamma -1}$$

I do not think there is a previous step that looks like the one you listed.

Edit: The original question statement is

Suppose 0.200 mol of an ideal diatmoic gas (y=1.4) undergoes a carnot cycle between 227 and 27 degrees celsius, starting at pressure A = 10x10^5 Pa at point a in the pV diagram (don't know how to post it). The volume doubles during isothermal expansion step a->b.

Is there anytihn you can take from this that might imply something?

Well, like I said, the derivation is relating the states to each other. That it is a Carnot cycle is additional info - do you know what that is?

You also know it is an ideal, diatomic, gas so you know the state equation.
Hence you have the entire state of point a.

Can you relate your PV diagram to this one

... I figure your point a is the point A above.

 

[RTW69]

This equation is derived from the ideal gas law and the first law of thermodynamics for Adiabatic (q=0) Reversible process, with constant c_v. If you Google it you can find the actual derivation if you like. It going to look something like:

du=-dw_rev=-pdv

c_vdT=(RT/V)dv and then calculus happens