Question

Mulțumesc anticipat!

Answer 1

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

$\displaystyle{ \nu = 2 \ moli }$

$\displaystyle{ \mu }$ = 28 kg/kmol = $\displaystyle{ 28 \cdot 10^{-3} }$ kg/mol

t₁ = 27°C ⇒ T₁ = 27 + 273 = 300 K

---------------------------------------------------------

a) m = ?

b) $\rho$ = ?

c) p₁ = ?

d) T₂ = ?

(încălzire izocoră, p₂ = 2p₁

e) reprezentare grafică în coordonate (p, T)

----------------------------------------------------------

a) $\displaystyle{ \nu = \frac{m}{\mu} \rightarrow m = \nu \cdot \mu }$

$\displaystyle{ m = 2 \cdot 28 \cdot 10^{-3} = 56 \cdot 10^{-3} \ kg }$

b) $\displaystyle{ \rho = \frac{m}{V} = \frac{56 \cdot 10^{-3}}{10^{-2}} = 56 \cdot 10^{-1} = 5,6 \ kg/m^{3} }$

c) $\displaystyle{ p_{1} \cdot V = \nu \cdot R \cdot T_{1} \rightarrow p_{1} = \frac{\nu \cdot R \cdot T_{1}}{V} }$

$\displaystyle{ p_{1} = \frac{2 \cdot 8,31 \cdot 300 }{10^{-2}} = \frac{4986}{10^{-2}} }$

$\displaystyle{ p_{1} = 4986 \cdot 10^{2} = 4,98 \cdot 10^{5} \ Pa }$

d) transformare izocoră ⇒ volumul este constant

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

$\displaystyle{ T_{2} = \frac{2 \not p_{1} \cdot T_{1} }{\not p_{1}} = 2 \cdot T_{1} }$

T₂ = 2 × 300 = 600 K

e) Vezi imaginea atașată.