- #31
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Here is my entropy change analysis for the irreversible path suggested by @mfig. If we apply the first law of thermodynamics to his path, we obtain $$mC_v(T_2-T_1)=Q-P_2(V_2-V_1)$$where ##P_2=(M_P+M_L)g/A## is the final pressure. Applying the ideal gas law to the work term in the above equation then gives: $$mC_v(T_2-T_1)=Q-mRT_2(1-\frac{V_1}{V_2})\tag{1}$$
Even though the present irreversible path bears little resemblance to the reversible polytropic path analyzed previously, as pointed out in post #29, we can still use the polytropic parameterization in terms of the polytropic parameter n to establish the final state of the gas, such that the range of values for the single parameter n from 0 to ##-\infty## span all possible final states of the gas for which the final pressure and final volume are greater than their values in the initial state. With this in mind, combining Eqn. 2 of post #29 with Eqn. 1 of the present develop gives:$$mC_v(T_2-T_1)=Q-mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{2}$$From this, it follows that the total heat flow over the irreversible path is given by:$$Q=mC_v(T_2-T_1)+mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{3}$$Since, for this irreversible path, all the heat transfer at the boundary takes place at the final temperature ##T_2##, the integral of dQ divided by the boundary temperature (i.e., the boundary where the heat flow occurs) is just equal to ##Q/T_2##: $$\int{\frac{dQ}{T_{boundary}}}=\frac{Q}{T_2}$$$$=mC_v\left(1-\frac{T_1}{T_2}\right)+mR\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{4}$$Calculations show that, for all values of (1) the "final state parameter" n in the range 0 to ##-\infty##, (2) the heat capacity ratio parameter ##1<\gamma<1.67##, and (3) the overall temperature ratio ##\frac{T_2}{T_1}>1## the integral of ##dQ/T_{boundary}## calculated from Eqn. 4 for this irreversible process path is less than the entropy change between the initial and final states calculated from Eqn. 3 of post #29 for the 2-step reversible path (or the reversible polytropic path). For example, for n = -1, ##\gamma=1.4##, and ##\frac{T_2}{T_1}=2##, we find that ##dQ/T_{boundary}=0.617mC_v## while ##\Delta S =0.832mC_v##. This is consistent with the Clausius inequality, which represents a mathematical statement of the 2nd law of thermodynamics.
Even though the present irreversible path bears little resemblance to the reversible polytropic path analyzed previously, as pointed out in post #29, we can still use the polytropic parameterization in terms of the polytropic parameter n to establish the final state of the gas, such that the range of values for the single parameter n from 0 to ##-\infty## span all possible final states of the gas for which the final pressure and final volume are greater than their values in the initial state. With this in mind, combining Eqn. 2 of post #29 with Eqn. 1 of the present develop gives:$$mC_v(T_2-T_1)=Q-mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{2}$$From this, it follows that the total heat flow over the irreversible path is given by:$$Q=mC_v(T_2-T_1)+mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{3}$$Since, for this irreversible path, all the heat transfer at the boundary takes place at the final temperature ##T_2##, the integral of dQ divided by the boundary temperature (i.e., the boundary where the heat flow occurs) is just equal to ##Q/T_2##: $$\int{\frac{dQ}{T_{boundary}}}=\frac{Q}{T_2}$$$$=mC_v\left(1-\frac{T_1}{T_2}\right)+mR\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{4}$$Calculations show that, for all values of (1) the "final state parameter" n in the range 0 to ##-\infty##, (2) the heat capacity ratio parameter ##1<\gamma<1.67##, and (3) the overall temperature ratio ##\frac{T_2}{T_1}>1## the integral of ##dQ/T_{boundary}## calculated from Eqn. 4 for this irreversible process path is less than the entropy change between the initial and final states calculated from Eqn. 3 of post #29 for the 2-step reversible path (or the reversible polytropic path). For example, for n = -1, ##\gamma=1.4##, and ##\frac{T_2}{T_1}=2##, we find that ##dQ/T_{boundary}=0.617mC_v## while ##\Delta S =0.832mC_v##. This is consistent with the Clausius inequality, which represents a mathematical statement of the 2nd law of thermodynamics.