Answer:
[tex] V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}[/tex]
And replacing the info we have:
[tex]V_2 = \frac{960 mm Hg * 550 mL *423.15 K}{473.15 K*830 mm Hg}=568.920 mL=0.5689 L[/tex]
Step-by-step explanation:
For this case we wan assume that we have an ideal gas and we can use this relation:
[tex]\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}[/tex]
And for this case we know:
[tex] P_1 = 960 mm Hg[/tex]
[tex] T_1 = 200+273.15 K =473.15 K[/tex]
Becuase [tex] K = C +273.15[/tex]
[tex] V_1 =550 ml = 0.55L[/tex]
Because 1L = 1000 mL
[tex]P_2 = 830 mm Hg[/tex]
[tex] T_2 = 150+273.15 K= 423.15K [/tex]
And if we solve for V2 we got:
[tex] V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}[/tex]
And replacing the info we have:
[tex]V_2 = \frac{960 mm Hg * 550 mL *423.15 K}{473.15 K*830 mm Hg}=568.920 mL=0.5689 L[/tex]