gaz diatomic ⇒ 5 grade de libertate ⇒ Cv = 5R/2, Cp = 7R/2
[tex]\displaystyle{ V = 2l = 2 \cdot 10^{-3} \ m^{3} }[/tex]
t₁ = 27°C ⇒ T₁ = 27 + 273 = 300K
p₁ = 100 kPa = 100.000 Pa = [tex]\displaystyle{ 10^{5} \ Pa }[/tex]
1 - 2 = transformare izocoră (⇒ V = ct)
p₂ = 300 kPa = 300.000 Pa = [tex]\displaystyle{ 3 \cdot 10^{5} \ Pa }[/tex]
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a) T₂ = ?
b) Q = ?
c) L = ?
d) ΔU = ?
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a)
1 - 2 = transformare izocoră ⇒ volumul rămâne constant
[tex]\displaystyle{ \frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}} \rightarrow T_{2} \cdot p_{1} = p_{2} \cdot T_{1} }[/tex]
[tex]\displaystyle{ T_{2} = \frac{p_{2} \cdot T_{1} }{p_{1}} }[/tex]
[tex]\displaystyle{ T_{2} = \frac{3 \cdot 10^{5} \cdot 300}{10^{5}} }[/tex]
T₂ = 3 × 300 = 900K
b)
[tex]\displaystyle{ Q = \nu \cdot Cv \cdot \Delta T = \nu \cdot \frac{5}{2} R \cdot (T_{2} - T_{1}) }[/tex]
[tex]\displaystyle{ p \cdot V = \nu \cdot R \cdot T \rightarrow Q = \frac{5}{2} \cdot (p_{2}V_{2} -p_{1}V_{1}) }[/tex]
[tex]\displaystyle{Q=\frac{ 5 \cdot (3 \cdot 10^{5} \cdot 2 \cdot 10^{-3} - 10^{5} \cdot 2\cdot 10^{-3}) }{2} }[/tex]
[tex]\displaystyle{ Q = \frac{5 \cdot 2 \cdot 10^{5} \cdot 2 \cdot 10^{-3}}{2} }[/tex]
Q = 5 × 2 × 100
Q = 1000 J
c)
L = 0
(in transformarile izocore nu se efectueaza lucru mecanic)
d)
ΔU = Q = 1000 J
Daca este ceva ce nu ai inteles ma poti intreba in comentarii.
