Napon pare

Submitted by pmf93 on Sun, 11/17/2013 - 17:44
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Opšta diskusija

Trebala bi mi pomoc oko jednog zadatka.

Napon pare HNO3 iznosi 27731 Pa na 50 C i 170919 Pa na 100 C. Kolika je: a) normalna temperatura kljucanja i b) molarna entalpija isparavanja azotne kiseline?

Kada uvrstim podatke u fomulu za izracunavanje temperature,

lnp2/p1=H/R * T2-T1/T2*T1

0.523=36464/8.314*373.15-T1/373.15*T1

kako da izrazim T1 iz ove formule?

 

 

milicaradenkovic's picture

milicaradenkovic - Sun, 11/17/2013 - 18:14

lnp2/p1=H/R*((T2-T1)/T2*T1)

0.523=36464/8.314*((373.15-T1)/373.15*T1

0.523*373.15*T1=4385.85*(373.15-T1)

195.16*T1=1.64*10(6)-4385.85*T1

4581.01*T1=1.64*10(6)

T1=358 K

pmf93's picture

pmf93 - Sun, 11/17/2013 - 18:22

Hvala.