The volume of a gas is 550 mL at 960 mm Hg and 200.0 o C. What volume would the

pressure of the gas be 830 mm Hg if the temperature is reduced to 150.0 o C?

Respuesta :

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]

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