Problem:
One kilomole of an ideal, monatomic gas undergoes a reversible Carnot-processes between temperatures 300 °C and 20 °C. The work done during one cycle is 1500 kJ.
a) Find the entropy-change in every process and show that total sum of entropy-changes is zero
b) What is the ratio between the largest and smallest volume that the gas takes on during the whole process?
I have already done a), it is b) I am struggling with. I realized that the greatest and smallest volume is assumed in the adiabatic processes of the Carnot-cycle, so I applied $T_1V_1^{\gamma -1} = T_1V_1^{\gamma -1}$ but to no avail. How can one do it?
Best Answer
The total work done by the system in one cycle is the sum of the works done in the isothermal processes:
$$W_{total}=nR\left ( T_h\ln \frac{V_2}{V_1}-T_c\ln\frac{V_3}{V_4} \right )$$
Also for adiabatic processes we have :
$$T_h V_2^{\gamma-1 }=T_c V_3^{\gamma-1}\tag{2}$$
$$T_hV_1^{\gamma-1}=T_cV_4^{\gamma-1}\tag{3}$$
Substituting for $V_2$ and $V_4$ from $(2)$ and $(3)$ in the first equation, we can find the ratio $V_3\over V_1$. (Find the final result yourself!)