Question

Daca are cine sa ma ajute cu aceasta problema si sa imi explice rezolvarea i-as fi recunoscator.
Multumesc!​

Answer 1

gaz diatomic ⇒ 5 grade de libertate ⇒ Cv = 5R/2, Cp = 7R/2

$\displaystyle{ V = 2l = 2 \cdot 10^{-3} \ m^{3} }$

t₁ = 27°C ⇒ T₁ = 27 + 273 = 300K

p₁ = 100 kPa = 100.000 Pa = $\displaystyle{ 10^{5} \ Pa }$

1 - 2 = transformare izocoră (⇒ V = ct)

p₂ = 300 kPa = 300.000 Pa = $\displaystyle{ 3 \cdot 10^{5} \ Pa }$

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a) T₂ = ?

b) Q = ?

c) L = ?

d) ΔU = ?

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a)

1 - 2 = transformare izocoră ⇒ volumul rămâne constant

$\displaystyle{ \frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}} \rightarrow T_{2} \cdot p_{1} = p_{2} \cdot T_{1} }$

$\displaystyle{ T_{2} = \frac{p_{2} \cdot T_{1} }{p_{1}} }$

$\displaystyle{ T_{2} = \frac{3 \cdot 10^{5} \cdot 300}{10^{5}} }$

T₂ = 3 × 300 = 900K

b)

$\displaystyle{ Q = \nu \cdot Cv \cdot \Delta T = \nu \cdot \frac{5}{2} R \cdot (T_{2} - T_{1}) }$

$\displaystyle{ p \cdot V = \nu \cdot R \cdot T \rightarrow Q = \frac{5}{2} \cdot (p_{2}V_{2} -p_{1}V_{1}) }$

$\displaystyle{Q=\frac{ 5 \cdot (3 \cdot 10^{5} \cdot 2 \cdot 10^{-3} - 10^{5} \cdot 2\cdot 10^{-3}) }{2} }$

$\displaystyle{ Q = \frac{5 \cdot 2 \cdot 10^{5} \cdot 2 \cdot 10^{-3}}{2} }$

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.