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Gases Unit 2. Combined Gas Law Combines the laws of Boyle (P,V), Charles (V, T), and Gay-Lussac (P,...
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Transcript of Gases Unit 2. Combined Gas Law Combines the laws of Boyle (P,V), Charles (V, T), and Gay-Lussac (P,...
Gases
Unit 2
Combined Gas Law
Combines the laws of Boyle (P,V), Charles (V, T), and Gay-Lussac (P, T)
Relates: P, V, Twhen n is constant
Equation: P1V1 = P2V2
T1 T2
Combined Gas Law
Has 3 variables and can also be used as any of the above minor gas laws
Suggestion:
Organize your info before you start!
Combined Gas Law
Examples:
A gas has a volume of 800.0 mL at 23.00oC and 300.0 torr. What would the volume of the gas be at 227.0oC and 600.0 torr of pressure?
Combined Gas Law
V1 = 800.0 mL
T1 = 23.00oC = 296 K
P1 = 300.0 torr
V2 = ? mL
T2 = 227.0oC = 500 K
P2 = 600.0 torr
Combined Gas Law
P1V1 = P2V2
T1 T2
(300.0 torr)(800.0 mL) = (600.0 torr)(V2) 296 K 500 K810.811 = 1.20(V2) 1.20 1.20V2 = 675.7 mL
Combined Gas LawA gas sample occupies 3.25 L at 24.5oC
and 350 kPa. Determine the temperature (in Celsius) at which the gas will occupy 4.25 L at 150 kPa.
V1 = 3.25 LT1 = 24.5oC = 297.5 KP1 = 350 kPaV2 = 4.25 LT2 = ? oCP2 = 150 kPa
Combined Gas Law
P1V1 = P2V2
T1 T2
(350 kPa)(3.25 L) = (150 kPa)(4.25 L)297.5 K T2
3.82 = 637.5 1 T2
Cross multiply…
Combined Gas Law
(3.82)(T2) = 637.5
3.82 3.82
T2 = 167 K
-273
T2 = -106oC
Combined Gas Law0.73 L of nitrogen at STP is heated to
80.0oC and the volume is increased to 4.53 L. What is the new pressure in mm Hg?
V1 = 0.73 L
T1 = 273 K
P1 = 760 mm Hg
V2 = 4.53 L
T2 = 80.0oC = 353 K
P2 = ? mm Hg
Combined Gas LawP1V1 = P2V2
T1 T2
(760 mm Hg)(0.73 L) = P2(4.53 L)
273 K 353 K2.03 = 0.0128(P2)
.0128 .0128P2 = 160 mm Hg
Combined Gas LawIf a gas in a rigid container has a pressure of
1.75 atm at standard temperature, what is the new pressure when it is heated to 50.0oC?
P1 = 1.75 atm
T1 = 273 K
P2 = ? atm
T2 = 50.0oC = 323 K
V1 = V2
Combined Gas Law
P1V1 = P2V2
T1 T2
(1.75 atm)(V) = (P2)(V) 273 K 323 K565.25(V) = (273)(P2)(V)273(V) 273(V)
P2 = 2.07 atm
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