Question

Intr un vas se gaseste o masa m=2,9 kg de aer la temperatura t1=300K si presiunea P=2×10la a 5a N/m la2. Se incalzeste masa de aer izobar pana la temperatura T2=600K
a) volumul intitial
b) ccantitatea de caldura
c) lucrul mecanic
d ) variatia energiei interne
Se da : Cp =7/2 R
M= 29 Kg Kmol ​

Answer 1

Răspuns:

a)

$1,2471\,m^3$

b)

$872970\,J$

c)

$249420\,J$

d)

$623550\,J$

Explicație:

a)

Cantitatea de moli:

$\nu = \dfrac{m}{\mu}=\dfrac{2,9\,kg}{29\,kg/kmol}=0,1\,kmoli = 100\,moli$

Legea gazelor ideale: (pentru starea 1, stare initiala)

$p V_1=\nu R T_1 \implies V_1=\dfrac{\nu R T_1 }{p}$

$V_1=\dfrac{100\cdot 8,314\cdot300}{2\cdot 10^5}=1,2471\,m^3$

b)

La transformarea izobara, caldura are urmatoarea formula:

$Q=\nu C_p\Delta T =\nu C_p(T_2-T_1)$

$Q=\nu \cdot \dfrac{7}{2}R(T_2-T_1)$

$Q=100\cdot\dfrac{7}{2}\cdot8,314\cdot(600-300)=872970\,J$

c)

Pentru transformarea izobara, avem legea Gay-Lussac:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \implies V_2=\dfrac{V_1}{T_1}T_2$

$V_2=\dfrac{1,2471}{300}\cdot600=2,4942\,m^3$

$L=p\Delta V=p(V_2-V_1)$

$L=2\cdot10^5\cdot(2,4942-1,2471)=249420\,J$

d)

Din principiul intai al termodinamicii:

$\Delta U=Q-L$

$\Delta U = 872970-249420=623550\,J$