Question

Va rog frumos, vreau un raspuns bun si corect!!!
Multumesc

Answer 1

[tex]\displaystyle \text{Se da:}\\ \\ V_2=100\times V_1\\ \\ T_1=10\times T_2\\ \\ C_v=?\\ \\ \\ \text{Formule:}\\ \\ \text{Avem o transformare adiabatica, deci din legea lui Poisson:}\\ \\ \\ p_1\times V_1^{\gamma}=p_2\times V_2^{\gamma}\\ \\ \\ \text{Unde }\gamma \text{ este exponentul adiabatic}\\ \\ \\ \text{Folosind formula:}\\ \\ \\ \frac{p_1\times V_1}{T_1}=\frac{p_2\times V_2}{T_2}\\ \\ \\ \text{Obtinem o alta forma a legii:}\\ \\ \\ T_1\times V_1^{\gamma-1}=T_2\times V_2^{\gamma-1}\\ \\ \\ [/tex]


[tex]\displaystyle \text{Folosind datele problemei, avem:}\\ \\ \\ 10\times T_2\times V_1^{\gamma-1}=T_2\times (100\times V_1)^{\gamma-1}\\ \\ 10\times V_1^{\gamma-1}=100^{\gamma-1}\times V_1^{\gamma-1}\\ \\ 10=100^{\gamma-1}\\ \\ 10=(10^{2)^{\gamma-1}}\\ \\ 10^1=10^{2\times \gamma-2}\\ \\ 1=2\times\gamma-2\\ \\ 2\times\gamma=3\\ \\ \gamma=1,5\\ \\ \\[/tex]


[tex]\displaystyle\text{Caldura molara izocora o vom afla din relatia lui Mayer:}\\ \\ \\ C_p-C_v=R\\ \\ \\ \text{Folosind coeficientul adiabatic, putem afla }C_p\text{ folosind formula:}\\ \\ \\ \frac{C_p}{C_v}=\gamma\Rightarrow C_p=C_v\times\gamma\\ \\ \\ C_v\times\gamma-C_v=R\\ \\ C_v\times (\gamma-1)=R\\ \\ C_v=\frac{R}{\gamma-1}\\ \\ \\ \text{Pentru cazul nostru, }\gamma=1,5\text{ avem:}\\ \\ \\ [/tex]

[tex]\displaystyle C_v=\frac{R}{1,5-1}\\ \\ C_v=\frac{R}{0,5}\\ \\ \\ \boxed{C_v=2\times R}[/tex]