Răspuns:
a)
[tex]1,2471\,m^3[/tex]
b)
[tex]872970\,J[/tex]
c)
[tex]249420\,J[/tex]
d)
[tex]623550\,J[/tex]
Explicație:
a)
Cantitatea de moli:
[tex]\nu = \dfrac{m}{\mu}=\dfrac{2,9\,kg}{29\,kg/kmol}=0,1\,kmoli = 100\,moli[/tex]
Legea gazelor ideale: (pentru starea 1, stare initiala)
[tex]p V_1=\nu R T_1 \implies V_1=\dfrac{\nu R T_1 }{p}[/tex]
[tex]V_1=\dfrac{100\cdot 8,314\cdot300}{2\cdot 10^5}=1,2471\,m^3[/tex]
b)
La transformarea izobara, caldura are urmatoarea formula:
[tex]Q=\nu C_p\Delta T =\nu C_p(T_2-T_1)[/tex]
[tex]Q=\nu \cdot \dfrac{7}{2}R(T_2-T_1)[/tex]
[tex]Q=100\cdot\dfrac{7}{2}\cdot8,314\cdot(600-300)=872970\,J[/tex]
c)
Pentru transformarea izobara, avem legea Gay-Lussac:
[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \implies V_2=\dfrac{V_1}{T_1}T_2[/tex]
[tex]V_2=\dfrac{1,2471}{300}\cdot600=2,4942\,m^3[/tex]
[tex]L=p\Delta V=p(V_2-V_1)[/tex]
[tex]L=2\cdot10^5\cdot(2,4942-1,2471)=249420\,J[/tex]
d)
Din principiul intai al termodinamicii:
[tex]\Delta U=Q-L[/tex]
[tex]\Delta U = 872970-249420=623550\,J[/tex]
