V = 10 l = [tex]\displaystyle{ 10 \cdot 10^{-3} = 10^{-2} \ m^{3} }[/tex]
[tex]\displaystyle{ \nu = 2 \ moli }[/tex]
[tex]\displaystyle{ \mu }[/tex] = 28 kg/kmol = [tex]\displaystyle{ 28 \cdot 10^{-3} }[/tex] kg/mol
t₁ = 27°C ⇒ T₁ = 27 + 273 = 300 K
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a) m = ?
b) [tex]\rho[/tex] = ?
c) p₁ = ?
d) T₂ = ?
(încălzire izocoră, p₂ = 2p₁
e) reprezentare grafică în coordonate (p, T)
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a) [tex]\displaystyle{ \nu = \frac{m}{\mu} \rightarrow m = \nu \cdot \mu }[/tex]
[tex]\displaystyle{ m = 2 \cdot 28 \cdot 10^{-3} = 56 \cdot 10^{-3} \ kg }[/tex]
b) [tex]\displaystyle{ \rho = \frac{m}{V} = \frac{56 \cdot 10^{-3}}{10^{-2}} = 56 \cdot 10^{-1} = 5,6 \ kg/m^{3} }[/tex]
c) [tex]\displaystyle{ p_{1} \cdot V = \nu \cdot R \cdot T_{1} \rightarrow p_{1} = \frac{\nu \cdot R \cdot T_{1}}{V} }[/tex]
[tex]\displaystyle{ p_{1} = \frac{2 \cdot 8,31 \cdot 300 }{10^{-2}} = \frac{4986}{10^{-2}} }[/tex]
[tex]\displaystyle{ p_{1} = 4986 \cdot 10^{2} = 4,98 \cdot 10^{5} \ Pa }[/tex]
d) transformare izocoră ⇒ volumul este constant
[tex]\displaystyle{ \rightarrow \frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}} \rightarrow T_{2} = \frac{p_{2} \cdot T_{1}}{p_{1}} }[/tex]
[tex]\displaystyle{ T_{2} = \frac{2 \not p_{1} \cdot T_{1} }{\not p_{1}} = 2 \cdot T_{1} }[/tex]
T₂ = 2 × 300 = 600 K
e) Vezi imaginea atașată.
