Question

Termodinamica
Fisica

Answer 1

a)

$EC = \frac{3}{2} .nR.T$

Ec = mv² / 2

fica :)

$\frac{m.v^2}{2} = \frac{3}{2} .nR.T$

$m.v^2 = 3.nR.T$

$V= \sqrt{\frac{ 3.nR.T }{m} }$

$V= \sqrt{\frac{ 3.nR.T }{m} }$

R = 8,31 J / mol;K

265°C = 538°K

0,1 Kg = 100 g de CO2  = 2,27 mol

$V= \sqrt{\frac{ 3.(2,27)(8,31).(538) }{0,1} } = 552,11~ m/s$

b)
$P.V = nR.T$

$P.(0,05) = (2,27)(8,31 } ).(538)$

$P = 202973,412 \frac{N}{m^2 } ~ ou ~ 2,029~ atm$

c) 
   $EC = \frac{3}{2} .2,27.(8,31).538 = 15223,00~ J$

D)  1 mol -------- 6.10²³ moléculas
     2,27 mol------x

x = 1,362 .10²⁴ moléculas