UPDATED - ALL QUESTION ARE ANSWERED
But I'm not sure with my question in effusion rate problems and I don't know how to make the analysis (#4 under Partial Pressure and Graham's Law)
But I'm not sure with my question in effusion rate problems and I don't know how to make the analysis (#4 under Partial Pressure and Graham's Law)
More Problems
1. Given:
V1 = 15 L
T1 = 67oC
+ 273 = 340 K
P1 = 107
kpa
T2 = 95oC
+ 273 = 368 K
P2 = 112
kPa
Required : V2
Calculation:
P1 x V1/T1
= P2 x V2/T2
107 kPa x 15 L/340 K = 112 kPa x V2/368 K
4.72 = 112 kPa x V2/368 K
112 kPa x V2 = 4.72 x 368 K
V2 = 1736.96 K/112 kPa
V2 = 15.5 L
Paraphrase: The volume of the new volume of a 15 L gas at 67oC
and 107 kPa when heated to 95oC and has 112 kPa is 15.5 L.
Evaluation: The increase of pressure and temperature makes the volume
increases too. The new volume is bigger than the previous volume, therefore it
makes sense.
2. Given:
V1 = 10
mL/1000 = 0.01 L
T1 = 42oC
+ 273 = 297 K
T2 = 283oC
+ 273 = 556 K
Required : V2
Calculation:
V1/T1
= V2/T2
0.01 L/297 K = V2/556
K
V2 = 0.01 L/297 K x 556 K
V2 = 0.01872 L x 1000 = 18.72 mL
Paraphrase: The new volume of a 10 mL container at 42oC when
the temperature is adjust to 283oC is 18.72 mL.
Evaluation: The increase of temperature increases the volume. The new
volume is bigger than the last volume, therefore it makes sense.
3. Given:
T1 = 25oC
+ 273 = 298 K
P1 = 101
kPa
P2 = 200
kPa
Required : T2
Calculation:
P1/T1
= P2/T2
101 kPa/298 K = 200
kPa/T2
T2(101
kPa/298 K) = 200 kPa
T2 = 200
kPa/(101 kPa/298 K)
T2 = 590 K
Paraphrase: The new temperature of a container of gas at 25oC
and 101 kPa is 590 K when the pressure increases to 200 kPa.
Evaluation: The increase of pressure increases the temperature. The new
temperature is bigger than the last temperature, so the answer makes sense.
4. Given:
V1 = 50 L
P1 = 101.3 kPa
P2 = 2.5 atm x 101.3 kPa = 253.25 kPa
Required: V2
Calculation:
P1 x V1
= P2 x V2
101.3 kPa x 50 L =
253.25 kPa x V2
V2 = (101.3
kPa x 50 L)/253.25 kPa
V2 = 20 L
Paraphrase : The
new volume of a 5 ml container at standard pressure is 20 L when the pressure
is extended to 2.5 atm.
Evaluation : The
increase of pressure results the volume to decrease. The new volume is smaller
than the previous volume, therefore the answer makes sense.
5. Given:
V1 = 10.5 L
T1 = 75oC
+ 273 = 348 K
P1 = 101
kPa
T2 = 125oC
+ 273 = 398 K
V2 = 11.5 L
Required : P2
Calculation:
P1 x V1/T1
= P2 x V2/T2
(101 kPa x 10.5 L)/348 K = (P2 x 11.5 L)/298 K
P2 = ([(101 kPa x 10.5 L)/384 L] x 298 K)/11.5 L
P2 = 71.56 kPa
Paraphrase: The new pressure of
10.5 L sample of gas at 75oC and 101 kPa is 71.56 kPa when is heated
to 125oC.
Evaluation: The increase in temperature and volume results the pressure
to decrease. The new pressure is smaller than the last pressure, the answer
should be right.
6. If the volume and mass of gases is different, the density is
different even though the gases have the same pressure and pressure. Gases with
same amount of mass but less in volume or gases that have the same volume but
have more mass will be denser than the other gases.
Ideal Gas Law Practice
1. Given:
V = 120 L
P = 202.3 kPa
T = 340 K
Required : n
Calculation:
n = PV/RT
n = (202.3 kPa x 120 L)/(8.31 LkPa/(Kmol) x 340 K)
n = 24276 LkPa/(2825.4
LkPa/mol)
n = 8.59 mol
Paraphrase : 120
L of gas at a pressure of 202.3 kPa and a temperature of 340 K has 8.59 of
moles.
Evaluation: Even
though the temperature is high, the pressure and the mass is big, therefore it
makes sens if the number of moles is quite a lot.
2. Given:
V = 50 L
n = 45 mol
T = 200oC + 273 = 473 K
Required: P
Calculation:
P = nRT/V
P = (45 mol x (8.31 Lkpa/Kmol) x 473 K)/50 L
P = 3537.57 kPa
Paraphrase: A 50 liter container that holds 45 moles of gas at a
temperature of 200oC has a pressure at 3537.57 kPa.
Evaluation: The number of moles and the temperature is high, so it
makes sense that the pressure is high.
3. Given:
V = 1 L
n = 2 mol
T = 1400oC
+ 273 = 1673 K
Required : P
Calculation:
P = nRT/V
P = (2 mol x (8.31 LkPa/Kmol) x 1673 K)/1 L
P = 27805.26 kPa
Paraphrase: The pressure inside the canister is 27805.26 kPa.
Evaluation: The temperature is really high while the mass is really small,
thus it is reasonable that the pressure is really high.
4. Given:
V = 30 L
T = 300 K
P = 2026 kPa
Required: n
Calculation:
n = PV/RT
n = (2026 kPa x 30 L)/((8.31 LkPa/Kmol) x 300 K)
n = 60780 LkPa/(2493 LkPa/Kmol)
n = 24.38 mol
Paraphrase: The number of moles
of gas in a 30 liter scuba canister with a temperature of 300 K and a pressure
of 2026 kPa is 24.38 moles.
Evaluation: Even though the temperature is quite high, the pressure is
really high and the volume also quite high, so it makes sense that the number
of moles is a lot.
5. Given:
V = 100 L
n = 3 mol
P = 101.3 kPa
Required: T
Calculation:
T = PV/nR
T = (101.3 kPa x 100
L)/(3 mol x (8.31 LkPa/Kmol))
T = 10130 LkPa/(24.93
LkPa/K)
T = 406.34 K
Paraphrase: The temperature
of the balloon is 406.34 K.
Analysis: The moles is
not too many while the mass is large, therefore it’s reasonable that the
temperature is also high.
6. Given:
V = 67 L
M = 5 g N
T = 25oC + 273 = 298 K
Required: P
Calculation:
nN = 5 g N x 1 mol N/14.0067 g N = 0.36 mol N
P = nRT/V
P = (0.36 mol N x (8.31 LkPa/Kmol) x 298 K)/67 L
P = 13.19 kPa
Paraphrase: 67 L container that holds 5 g of nitrogen at a temperature
of 25oC has a pressure of 13.19 kPa.
Analysis: Even though the volume is large, the temperature is high and
the moles isn’t too small, therefore it’s reasonable that the pressure is
small.
7. Given:
V = 32 L
nO = 5.4 x 1023 mol O
P = 101.3 kPa
Required: T
Calculation:
T = PV/nR
T = (101.3 kPa x 32
L)/(5.4 x 1023 mol x (8.31 LkPa/Kmol))
T = 3241.6 LkPa/(4.4874
x 1024 LkPa/K)
T = 7.22 x 10-22 K
Paraphrase: The temperature inside a 32 L container that holds 5.4 x 1023
molecules of oxygen at a pressure of 101.3 kPa is 7.22 x 10-22 K.
Analysis: The moles are really high in amount, so it makes sense that
the temperature is really small.
Partial Pressure and Graham’s law Practice
1. Given:
PO2 = 3 atm x 101.3 = 303.9 kPa
PN2 = 5 atm x 101.3 = 506.5 kPa
PHe = 7 atm x 101.3 = 709.1 kPa
Required: Ptotal
Calculation:
Ptotal = PO2 + PN2 + PHe
Ptotal = (303.9 + 506.5 + 709.1) kPa
Ptotal = 1519.5 kPa
Paraphrase: The total pressure inside the tank is 1519.5 kPa.
Analysis: The total pressure is higher than the partial pressures,
therefore it makes sense because the partial pressure adds up to get the total
pressure.
2. Given:
Ptotal = 1.45 atm x 101.3 = 146.885 kPa
PCO2 = 0.43 atm x 101.3 = 43.559 kPa
PH2S = 0.23 atm x 101.3 = 23.299 kPa
Required: P3
Calculation:
P3 = Ptotal – (PCO2 – PH2S)
P3 = 146.885 kPa – (43.559 kPa + 23.299 kPa)
P3 = 80.027 kPa
Paraphrase: The partial pressure of the remaining air is 80.027 kPa.
Analysis: The partial pressure of the remaining air is a little bit
more than a half of the total pressure, it makes sense because the other
partial pressure are a little bit fewer than the total pressure if added up.
3. Given:
Ptotal = 1.45 atm x 101.3 = 146.885 kPa
PCO2 = 0.43 atm x 101.3 = 43.559 kPa
PH2S = 0.23 atm x 101.3 = 23.299 kPa
P3 = 80.027 kPa
Required: PO2
Calculation:
PO2 = 0.22 x P3
PO2 = 0.22 x 80.027 kPa
PO2 = 17.606 kPa
Paraphrase: The partial pressure
of oxygen in the air is 17,606 kPa
Analysis: The oxygen is the part of the air, so it makes sense that the
pressure is smaller than the air pressure.
4. Given:
Molar mass of O2 = 32.0 g/mol
Molar mass of N2 = 28.0 g/mol
Required: O2/N2
Calculation: O2/N2 = √(28.0 g/mol
/ 32.0 g/mol)
O2/N2 = 0.9354
Paraphrase: The rate of effusion of O2 and N2 is
0.9354
5. Given:
V = 27 L
T = 298 K
nN2 = 3
nO2 = 4
Required: P
Calculation:
P1 = (nN2)RT/V
P1 = (3 mol x 8.31 LkPa/Kmol x 298 K)/27 L
P1 = 2429.14 LkPa/27 L
P1 = 275.15 kPa
P2 = (nO2)RT/V
P2 = (4 mol x 8.31 LkPa/Kmol x x298 K)/27 L
P2 = 9905.52 LkPa/27 L
P2 = 366.87 kPa
Ptotal = (275.15 + 366.87) kPa
Ptotal = 642.02 kPa
Paraphrase: The pressure of the resulting gases is 642.02 kPa.
Analysis: The answer makes sense
because both the values of the volume and temperature are high, therefore the
pressure isn’t too high or too low.
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