Question

Se amesteca 2 gaze, unul monoatomic si unul biatomic astfel încât cantitatea de gaz monoatomic reprezinta o fractiune f din cantitatea totală.
Care este căldură molara izocora a amestecului?
Ce valoarea ar trebui sa aiba procentul f astfel încât exponentul adiabatic (semnul acela ) este 1,5.

Answer 1

[tex]\displaystyle\text{Se da:}\\ \\ \gamma_1=\frac 53\\ \\ \gamma_2=\frac 75\\ \\ \nu_1=f\times\nu\\ \\ a)C_v=?\\ \\ b)\gamma'=1,5\text{; }f=?\\ \\ \\[/tex]


[tex]\text{Formule:}\\ \\ \text{Dupa amestecarea energia interna va fi suma precedentelor:}\\ \\ \\ \Delta U=\Delta U_1+\Delta U_2\\ \\ C_v\times\nu\times\Delta t=C_{v_1}\times\nu_1\times\Delta t+C_{v_2}\times\nu_2\times\Delta t_2\\ \\ C_v\times\nu=C_{v_1}\times\nu_1+C_{v_2}\times\nu_2\\ \\ \\ \text{Deoarece }\nu_1=f\times\nu,\text{ atunci }\nu_2=(1-f)\times\nu\\ \\ \\ C_v\times\nu=C_{v_1}\times f\times\nu+C_{v_2}\times(1-f)\times\nu\\ \\ C_v=C_{v_1}\times f+C_{v_2}\times(1-f)\\ \\[/tex]


[tex]\displaystyle \text{Folosind relatia pentru coeficientul adiabatic: } \gamma=\frac{C_p}{C_v}\\ \\ \text{Si relatia lui Mayer: }C_p-C_v=R\\ \\ \text{Obtinem expresia pentru caldura molara izocora: }C_v=\frac{R}{\gamma-1}\\ \\ \text{Astfel obtinem:}\\ \\ \\ C_v=\frac{R}{\gamma_1-1}\times f+\frac{R}{\gamma_2-1}\times(1-f)\\ \\ C_v=R\times(\frac{f}{\gamma_1-1}+\frac{1-f}{\gamma_2-1})\\ \\ C_v=R\times \frac{f\times\gamma_2-f+\gamma_1-\gamma_1\times f-1+f}{(\gamma_1-1)\times(\gamma_2-1)}\\ \\ \\ [/tex]


[tex]\displaystyle C_v=R\times \frac{f\times\gamma_2-f+\gamma_1-\gamma_1\times f-1+f}{(\gamma_1-1)\times(\gamma_2-1)}\\ \\ \\ \boxed{C_v=R\times \frac{f\times(\gamma_2-\gamma_1)+\gamma_1-1}{(\gamma_1-1)\times(\gamma_2-1)}}[/tex]


[tex]\displaystyle b)\text{Din formula exprimata de mai sus, avem:}\\ \\ \\ C_v'=R\times \frac{f\times(\gamma_2-\gamma_1)+\gamma_1-1}{(\gamma_1-1)\times(\gamma_2-1)}\\ \\ \\ \text{De unde:}\\ \\ \\ f\times(\gamma_2-\gamma_1)+\gamma_1-1=\frac{C_v'\times(\gamma_1-1)\times(\gamma_2-1)}{R}\\ \\ \\ \text{Unde }C_v'\text{ il gasim din formula exprimata de mai sus: }C_v=\frac{R}{\gamma-1}\\ \\ \\ f\times(\gamma_2-\gamma_1)+\gamma_1-1=\frac{(\gamma_1-1)\times(\gamma_2-1)}{\gamma-1}\\ \\ [/tex]

[tex]\displaystyle f\times(\gamma_2-\gamma_1)=\frac{\gamma_1\times\gamma_2-\gamma_1-\gamma_2+1}{\gamma-1}-\gamma_1+1\\ \\ f=\frac{\gamma_1\times\gamma_2-\gamma_1-\gamma_2+1-\gamma_1\times\gamma-\gamma+\gamma_1+1}{(\gamma-1)\times(\gamma_2-\gamma_1)}\\ \\ f=\frac{\gamma_2\times(\gamma_1-1)-\gamma\times(\gamma_1+1)+2}{(\gamma-1)\times(\gamma_2-\gamma_1)}\\ \\ \\[/tex]