Question

Intr un balon V=0,2 m cubi se afla heliu la P1=10^5 Pa si t=17 grade Celsius.Dupa ce in balon a mai fost introdus heliu, presiunea a devenit P2=3 ori 10^5 Pa iar temperatura t2=47 grade Celsius.Cu cat a crescut masa heliului din balon stiind ca miu=4 kg/mol?

Answer 1

Problema e formulata gresit la sfirsit, miu la geliu e 4kg/kmol.
In orice caz daca nu ma crezi, schimbi in calcule.

[tex]\displaystyle \text{Se da:}\\ \\ V=0,2m^3\\ \\ p_1=10^5Pa\\ \\ T_1=17^\circ=290K\\ \\ p_2=3\times 10^5Pa\\ \\ T_2=47^\circ=320K\\ \\ \mu=4\frac{kg}{kmol}=0,004\frac{kg}{mol}\\ \\ \Delta m=?kg\\ \\ \\[/tex]


[tex]\displaystyle Formule:\\ \\ \Delta m=m_2-m_1\\ \\ \\ \text{Avem de a face cu o transformare izocora:}\\ \\ \text{Din ecuatia Clapeyron-Mendeleev:}\\ \\ \\ p_1\times V=\frac{m_1}{\mu}\times R\times T_1\\ \\ m_1=\frac{p_1\times V\times\mu}{R\times T_1}\\ \\ \\ \text{Analogic pentru }m_2:\\ \\ \\ m_2=\frac{p_2\times V\times\mu}{R\times T_2}\\ \\ \\ \text{Inlocuim in expresia initiala:}\\ \\ \\ \Delta m=\frac{p_2\times V\times\mu}{R\times T_2}-\frac{p_1\times V\times\mu}{R\times T_1}\\ \\ \\[/tex]


[tex]\displaystyle \text{Scoatem factorul comun si obtinem formula finala:}\\ \\ \\ \boxed{\Delta m=\frac{V\times\mu}{R}\times(\frac{p_2}{T_2}-\frac{p_1}{T_1})}\\ \\ \\ \text{Calcule:}\\ \\ \Delta m=\frac{0,2\times0,004}{8,31}\times(\frac{3\times 10^5}{320}-\frac{10^5}{290})\approx 0,057kg[/tex]