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?

  • $\begingroup$ Did you convert temperature to degrees Kelvin? $\endgroup$ – David H Oct 27 '13 at 17:04
  • $\begingroup$ Sure. But the maximum and smallest volume lies on different adiabatic curves, so that's an issue I am not sure how to bypass. $\endgroup$ – thermophysics Oct 27 '13 at 17:17

The total work done by the system in one cycle is the sum of the works done in the isothermal processes: enter image description here $$W_{total}=nR\left ( T_h\ln \frac{V_2}{V_1}-T_c\ln\frac{V_3}{V_4} \right )$$ 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!)

  • $\begingroup$ What about work done in the adiabatic processes? Don't they contribute to the total work? $\endgroup$ – thermophysics Oct 27 '13 at 18:42
  • $\begingroup$ In the adiabatic processes $\Delta U=\Delta W$. Also, $\Delta U$ only depends on the temperature difference, so the total work in these two processes is zero. $\endgroup$ – Mo_ Oct 27 '13 at 19:24
  • $\begingroup$ Yes but the temperature is changing (decreasing in fact) and thus $\Delta U \neq 0$, right? Or wait do you mean that work done from 2 to 3 cancels out with the work done from 4 to 1? $\endgroup$ – thermophysics Oct 27 '13 at 19:31
  • $\begingroup$ Yeah, exactly . $\endgroup$ – Mo_ Oct 27 '13 at 19:34
  • $\begingroup$ All right thanks. By the way: The work done in the adiabatic process (from 2 to 3), is it $\frac {nRT_2 - nrT_1}{\gamma -1}$ where $T_2$ is the temperature for the isotherm from 3 to 4 and $T_1$ is the temperature for the isotherm from 1 to 2. $\endgroup$ – thermophysics Oct 27 '13 at 19:51

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