Question

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?

Answer 1

`V_2 = (P_1 V_1 T_2)/(T_1 P_2)`

And replacing the info we have:

`V_2 = (960 mm Hg * 550 mL *423.15 K)/(473.15 K*830 mm Hg)=568.920 mL=0.5689 L`

Explanation:

For this case we wan assume that we have an ideal gas and we can use this relation:

`(P_1 V_1)/(T_1) =(P_2 V_2)/(T_2)`

And for this case we know:

`P_1 = 960 mm Hg`

`T_1 = 200+273.15 K =473.15 K`

Becuase
`K = C +273.15`

`V_1 =550 ml = 0.55L`

Because 1L = 1000 mL

`P_2 = 830 mm Hg`

`T_2 = 150+273.15 K= 423.15K`

And if we solve for V2 we got:

`V_2 = (P_1 V_1 T_2)/(T_1 P_2)`

And replacing the info we have:

`V_2 = (960 mm Hg * 550 mL *423.15 K)/(473.15 K*830 mm Hg)=568.920 mL=0.5689 L`