# 340.05.01-03_Gases Instruments_Ideal_Combined_Dalton

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```Gases
KMT – Kinetic Molecular Theory of Gases
• Background for understand physical properties of
gases
• Assumptions:
o gases are mostly empty space
o molecules are in constant and chaotic motion
o collisions are complete elastic
o pressure caused by molecules colliding with
container walls (KE = &frac12; mv2)
• KMT can be used for all three states of matter.
Motion:
State
Gas
Liquid
Solid
translational
rotational
vibrational
✔
(✔)
✔
✔
✔
✔
✔
Measurements
Volume
Amount
Temperature
Pressure
1L
n = m / MM
oC + 273 = K
1 atm
760 mmHg (torr)
101.325 kPa
14.7 psi
(1.013 bar)
(1 cm3 = 1 mL)
(1 m)3 = (102 cm)3
Standard Temperature &amp; Pressure (STP) for Gas Pbs.
• T = 0oC (273 K)
P = 1 atm
Standard molar volume (@STP) = 22.4 L
Measurements – Measuring Devices
Barometer
(atmospheric)
Manometer
(a gas)
Sphygmomanometer
Measurements - Manometer
Pgas = Patm
Pgas = Patm
P gas &gt; Patm
Pgas = Patm + DHg
Pgas &lt; Patm
Pgas = Patm – DHg
Combined Gas Law
Combination of Four Gas Laws
• Boyle
(P &amp; V)
• Charles
(V &amp; T)
• Gay-Lussac (P&amp;T)
• Avogadro
PV = k1
V
= k2
T
P1V1 = P2V2
V1 V2
=
T1 T2
P
= k3
T
V
= k4
n
P1 P2
=
T1 T2
V1 V2
=
n1 n2
(at a fixed T &amp; P,
V is directly related to number of moles)
Graphically (Ideal Gases)
Boyle’s Law
Charles’s Law
Gay-Lussac’s Law
Combined Gas Law
P1V1 P2V2
=
n1T1 n2T2
A gas-filled balloon contains 1.1 mol in a 26.5 L at 25oC and 101
kPa of pressure. How many moles of gas are added or
removed to produce 27.0 L at 52oC and 1050 mmHg?
101kPa
1atm
P2V2 n1T1
P1 =
x
= 1.00atm
n2 =
101.325kPa
T2 P1V1
V1 = 26.5L
n1 = 1.1mol
(1.38atm)(27.0L)(1.1mol)(298K)
n2 =
T1 = 25 + 273 = 298K
(325K)(1.0atm)(26.1L)
1050mmHg
1atm
P2 =
x
= 1.38atm
760mmHg
n2 =1.4mol
V2 = 27.0L
Dn = +0.3mol
n2 = ?
T2 = 52 + 273 = 325K
Combined Gas Law
1. A deodorant can has a volume of 175 mL and a pressure of
3.8 atm at 22oC. What would the pressure be if the can was
heated to 100.oC?
P1 = 3.80atm
V1 = V2
n1 = n2
T1 = 22 + 273 = 295K
P2 = ?
T2 = 100 + 273 = 373K
P1V1 P2V2
=
n1T1 n2T2
P1 T2
P2 =
T1
(3.8atm)(373K)
P2 =
(295)
P2 = 4.8atm
IDEAL GAS LAW
PV = nRT
Units: P = atm
V= L
n = mol
R = 0.0821 Latm/ Kmol
T=K
PV = nRT
• A 47.3-L container containing 1.62 mol of He is heated until
the pressure reaches 1.85 atm. What is the temperature?
P = 1.85 atm
V = 47.3 L
n = 1.62 mol
R = 0.0821 L-atm/K-mol
T=?
T = (1.85 atm)(47.3 L)
(1.62 mol)(0.0821 L-atm/K-mol)
T = 658K (=385oC)
Ideal Gas Law – MOLAR MASS
rearrange for n: n = PV
RT
Let M = molar mass (g/mol) (m=mass of sample)
(n=number of moles)
M = m/n
=&gt;
M = mRT
PV
Ideal Gas Law – DENSITY
M = mRT
PV
M = m RT
V P
M = dRT
P
IDEAL GAS LAW
conventional form:
PV = nRT
molar mass
MM = mRT
PV
density
MM = dRT
P
Using Gas Laws to Determine a Balanced Equation
A piston expands after a gaseous reaction that is
carried out under constant pressure &amp; volume.
(Assume isolated system)
Which of the following balanced chemical equations
describes the reaction?
(1) A2(g) + B2(g)  2AB(g)
(3) A(g) + B2(g)  AB2(g)
(2) 2AB(g) + B2(g)  2AB2(g)
(4) 2AB2(g)  A2(g) + 2B2(g)
Using Gas Laws to Determine a Balanced Equation
1. Vinit &amp; Tinit (from picture)
2. Pinit = Pinit (from problem)
3. Combined gas law: P1 V1 = P2 V2
n1 T1 n2 T2
4. From Avogadro’s law,
if T doubles, n must be halved. (P &amp; V are constant)
number of moles
(1) A2(g) + B2(g) 
2AB(g)
(g) + B (g)  2AB (g)
(2) 2AB
2
2
1
+
1  2
2
+
1  2
(3) A(g) + B2(g)  AB2(g)
1
+
1  1
(4) 2AB2(g)  A2(g) + 2B2(g)
2
 1 +
2
Problem
Calculate the mass of sodium hydrogen carbonate necessary to
produce 2.87 L of carbon dioxide at 25oC and 2.00 atm.
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
1. How many moles of CO2(g) are produced?
PV
PV = nRT &THORN; n =
RT
(2.00atm)(2.97L)
n=
= 0.234 mol CO2
(0.0821L.atm / K.mol)(298K)
Problem
Calculate the mass of sodium hydrogen carbonate necessary to
produce 2.87 L of carbon dioxide at 25oC and 2.00 atm.
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
n = 0.234 mol CO2
2. Stoichiometry: 1 CO2 : 1 NaHCO3
3. Determine mass of 0.234 mole NaHCO3
0.234 mol NaHCO3 84.01 g NaHCO3
x
= 19.66 g mol NaHCO3
1 mol NaHCO3
Gas Mixtures (Dalton’s Law of Partial Pressures)
PT = Pi + Pii ...
mole fraction (Xi) = Pi/PT
so, Pi = Xi PT
When 1 mole of methane (CH4) is heated with 4 moles of O2,
1 mol CO2 and 2 mol H2O are formed. What are the partial
pressures if PT = 1.26 atm.
1CO2 + 2H2O  1CO2 + 2H2O
CO2: mole fraction:
partial pressure:
XCO2 = 1.00 / 5.00
0.200 * 1.26 atm
(all of the methane is consumed so n = 5)
= 0.200
= 0.252 atm
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