Question

um recipiente de 20l de capacidade contem certa massa de um gás submetida a 3 atm e 27°C. calcule o volume ocupado por essa massa quando submetida 2 atm E 47°c.

Answer 1

V1 = 20L

P1 = 3atm

T1 = 27°C = 300K

 

V2 = ?

P2 = 2atm

T2 = 47°C = 320K

 

$<var>\frac{P1*V1}{T1} = \frac{P2*V2}{T2}</var>$

 

$<var>\frac{3*20}{300} = \frac{2*V2}{320}</var>$

 

$<var>\frac{60}{300} = \frac{2V2}{320}</var>$

 

$<var>0,2 = \frac{2V2}{320}</var>$

 

$<var>64 = 2V2 </var>$

 

$<var>\frac{64}{2} = V2</var>$

 

$\<var>boxed{V2 = 32L}</var>$

 

Answer 2

$<var>V_{1} = 20L \\P_{1} = 3atm \\T_{1} = 27 + 273 = 300K \\ \\V_{2} =? \\P_{2} = 2atm \\T_{2} = 47 + 273 = 320K \\ \\\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}} \\\frac{3\cdot20}{300} = \frac{2V_{2}}{320} \\\frac{60}{300} = \frac{2V_{2}}{320} \\600V_{2}= 19200 \\V_{2} = \frac{19200}{600} \\V_{2} = 32L</var>$