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One mole of an ideal monoatomic gas is contained in a cylinder of volume $10_{−3}m_{3}$ at a temperature of $400K$. The gas is taken to a final state with a volume of $2×10_{−3}m_{3}$ and a temperature of $400K$ (note that this is an isothermal expansion process of the ideal gas). A thermal reservoir (cold source) with temperature Tres $=300K$ is available as well as a reversible working source. What is the maximum work that can be produced at FTR (reversible work source)? What is the thermodynamic efficiency of this process? Remember that for an ideal gas the fundamental equation for fixed $N$ can be written as, $S=cNRlnU+NRlnV+cte$ and that the thermal equation of state is given by $T1 =ucR $ Remember that energy conservation demands that $ΔU+Q_{FCR}+W_{FRT}=0$ and that the reversibility condition requires that $ΔS+ΔS_{FCR}=0$ Since FCR is a reservoir, it is worth that $QFCR=Q_{res.}$. Remember that for quasi-static processes when the FCR (reversible heat source) is a reservoir, the relation $Q_{FCR}=T_{FCR}ΔS_{FCR}=Q_{res}=T_{res}ΔS_{res}$ is also valid.

To find the maximum work that can be produced by the reversible work source and the thermodynamic efficiency of the process, we can use the given equations and information. Given: Initial volume, V? = 10^(-3) m³ Final volume, V? = 2 * 10^(-3) m³ Initial temperature, T? = 400 K Final temperature, T? = 400 K Reservoir temperature, T_res = 300 K We know that for an isothermal process of an ideal gas, the change in internal energy (?U) is zero because the temperature remains constant. So, the energy conservation equation (Equation 2.3) simplifies to: Q_FCR + W_FRT = 0 (since ?U = 0) Since the process is isothermal, we can use the ideal gas equation of state to express the pressures in ...