# Temperature Dependence of Thermal Efficiency in Carnot Cycle

Suppose a Carnot Cycle operates between a hot reservoir at temp $$T_h$$ and a cold reservoir at temp $$T_c$$. Suppose also that the system is initially at point $$(V_1,p_1)$$ in the pV-plane and that the subsequent states of the system in the pV-plane are $$(V_2,p_2)$$, $$(V_3,p_3)$$ and $$(V_4,p_4)$$, respectively. Finally, suppose also that the heat absorbed by the system during isothermal expansion is $$Q_{12}$$ and that the heat surrendered to the cold reservoir during isothermal compression is $$Q_{43}$$. Then, the thermal efficiency of the cycle is given by

$$$$\eta = 1 - \frac{Q_{43}}{Q_{12}} = 1 - \frac{T_c}{T_h} \quad (1)$$$$

However, by the Ideal Gas Law, the above can be written as

$$\eta = 1 - \frac{T_c\cdot \ln(V_3/V_4)}{T_h\cdot \ln(V_2/V_1)}$$

Now, for the second equality in (1) to hold, it must be the case that $$V_3/V_4 = V_2/V_1$$. How do I know apriori that this is true? There's no obvious reason I can discern that the above statement would necessarily be true.

• This is where we need it to be in a cycle. There are more relations. In essence, the relationship needed just means that the number of molecules in the gas is a constant. Commented Jun 27, 2023 at 5:00

Write your result as $$\eta = 1 - \frac{T_c \Delta S_{34}}{T_h\Delta S_{21}}$$ and recall that the Carnot cycle is reversible, so that $$\Delta S_{12}+\Delta S_{34}=0$$, which gives the correct efficiency.
You just have to equate the volumes through their laws in the cicle, two isothermal (1-2 & 3-4) and two adiabatic curves (2-3 & 4-1). In the first you have (i) $$p_1V_1 = p_2V_2$$ (iii) $$p_3 V_3 = p_4 V_4$$ Then for the adiabatic curves, (ii) $$p_2 V_2^\alpha = p_4 V_4^\alpha$$ (iv) $$p_2 V_2^\alpha = p_4 V_4^\alpha$$ in which $$\alpha$$ depends on the gas.
Now, working with eqs. (i) to (iv) you can demonstrate that your (1) is the same as your second equation, that is $$V_3/V_4=V_2/V_1$$.