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

Answer:

$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$

And replacing the info we have:

$V_2 = \frac{960 mm Hg * 550 mL *423.15 K}{473.15 K*830 mm Hg}=568.920 mL=0.5689 L$

Step-by-step explanation:

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

$\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}$

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:

$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$

And replacing the info we have:

$V_2 = \frac{960 mm Hg * 550 mL *423.15 K}{473.15 K*830 mm Hg}=568.920 mL=0.5689 L$