1.Convert 534 mmHg to atm, torr, and kPa (kilo Pascal) 2. A gas occupying a volu
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Question
1.Convert 534 mmHg to atm, torr, and kPa (kilo Pascal)
2. A gas occupying a volume of 575 mL at a pressure of 0.970 atm is allowed to expand at constant temperature until it pressure reaches 0.541 atm. What is its final volume (show your work) (Hint: Boyle’s Law)
3. A sample of a gas occupies 4.25 L at 125oC. At what temperature will the gas occupy a volume of 1.83 L if the pressure remains constant? (show your work) (Hint: Charles Law)
4. A balloon contains 0.128 mol of gas and has a volume of 2.76 L. If an additional 0.073 mol of gas is added to the balloon (at the same temperature and pressure) what will its final volume be(Hint: Avogadro’s Law) ? (show your work) Gas Constant R = 0.08206 L.atm/mol.K
5. A sample of a gas has a mass of 827 mg. Its volume is 0.27 L at a temperature of 88 degree Celsius and a pressure of 975 mm Hg. Find it’s molar mass (Ex 5.8) (show your work)
6. A sample of hydrogen gas is mixed with water vapor. The mix has a total pressure of 755 torr and the water vapor has a partial pressure of 24 torr. What amount ( in moles) of hydrogen gas is contained in 1.55 L of this mixture at 298 K? (Show your work. Look at Ex 5.9 on Page 188)
7. Look at Example 5.12 for the equation and answer the following What volume (in liters) of carbonmonoxide gas , measured at a temperature of 312 K and a pressure of 876 mm Hg, is required to synthesize 29.8 g of methanol.
8. How many liters of oxygen (at STP) are required to form 10.5 g of H2O ? Show your work (EX 5.13) 2 H2(g) + O2(g) 2 H2O(g)
Explanation / Answer
Question 1: Convert 534 mmHg to atm, torr, and kPa (kilo Pascal)
Answer:
760 mmHg = 1 atm = 760 torr = 101.325 kPa
a ) so 534 mmHg = 534/760 = 0.702 mmHg
b ) so 534 mmHg = 534 torr
c ) so 534 mmHg = 534 x 101.325 / 760 = 71.19 kPa
Question 2: A gas occupying a volume of 575 mL at a pressure of 0.970 atm is allowed to expand at constant temperature until it pressure reaches 0.541 atm. What is its final volume (show your work)
Answer:
Boyle’s Law states: P1V1 = P2V2
P1, Initial gas pressure = 0.970 atm
V1, Initial gas volume = 575 mL
P2, Final gas pressure = 0.541 atm
V2, Final gas volume = P1V1 / P2 = 0.970 x 575 / 0.541 = 1030.96 = 1031 mL = 1.03 L
Question 3: A sample of a gas occupies 4.25 L at 125oC. At what temperature will the gas occupy a volume of 1.83 L if the pressure remains constant? (show your work) (Hint: Charles Law)
Answer:
Charles Law states: T1V2 = T2V1
T1, Initial temperature = 125 °C + 273 K = 398 K
V1, Initial gas volume = 4.25 L
V2, Final gas volume = 1.83 L
T2, Final temperature = T1V2 / V1 = 398 x 1.83 / 4.25 = = 171 K - 273 = -102 oC
Question 4: A balloon contains 0.128 mol of gas and has a volume of 2.76 L. If an additional 0.073 mol of gas is added to the balloon (at the same temperature and pressure) what will its final volume be(Hint: Avogadro’s Law) ? (show your work) Gas Constant R = 0.08206 L.atm/mol.K
Answer:
Avogadro’s Law states: V1n2 = V2n1
V1, Initial gas volume = 2.76 L
n1, Initial number of moles = 0.128 mol
n2, Final number of moles = 0.128 + 0.073 = 0.201 mol
V2, Final gas volume = V1n2 / n1 = 2.76 x 0.201 / 0.128 = 4.33 L
Question 5. A sample of a gas has a mass of 827 mg. Its volume is 0.27 L at a temperature of 88 degree Celsius and a pressure of 975 mm Hg. Find it’s molar mass (show your work)
Answer:
Ideal law gas states: pV = nRT
Where n = number of moles of gas, R = the universal gas constant = 8.314 J mol-1 K-1
pressure of gas = 975 mm Hg = 130 kPa
volume of gas = 0.27 L
temperature of gas = 88 °C + 273 = 361 K
mass of gas = 827 mg = 0.827 g
According to ideal gas number of moles of a gas, n = pV / RT
number of moles of gas = 130 x 0.27 / 8.314 x 361 = 35.1 / 3001.35 = 0.0117 moles
molar mass of gas = mass of gas / number of moles of gas
molar mass of gas = 0.827 / 0.0117 = 70.68 g/mole = 71 g/mole
Question 6: A sample of hydrogen gas is mixed with water vapor. The mix has a total pressure of 755 torr and the water vapor has a partial pressure of 24 torr. What amount ( in moles) of hydrogen gas is contained in 1.55 L of this mixture at 298 K?
Answer:
Total pressure of mixture of hydrogen gas and water vapour = 755 torr = 100.66 kPa
water vapor partial pressure = 24 torr = 3.2 kPa
p, hydrogen gas partial pressure = 755 – 24 = 731 torr = 97.37 kPa
T, Temperature = 298 K
V, Volume of gas = 1.55 L
R, Gas constant = 8.314 J mol-1 K-1
number of moles of gas, n = pV / RT
= 97.37 x 1.55 / 8.314 x 298
= 150.92/ 2477.57
number of moles of gas, n = 0.061 moles
Question 7: What volume (in liters) of carbonmonoxide gas , measured at a temperature of 312 K and a pressure of 876 mm Hg, is required to synthesize 29.8 g of methanol.
Answer:
CO + 2H2 ………………..> CH3OH
Mass of Methanol to be synthesized = 29.8 g
From the above equation one can deduce that
32 g of of methanol is synthesized from 28 g of Carbon monoxide.
So 29.8 g mthanol can be synthesized from = 29.8 x 28 / 32 = 26.075 g of CO (0.931 moles)
n, number of moles of CO = 0.931 moles
T, Temperature = 312 K
P, Pressure = 876 mmHg = 116.79 kPa
R, Gas constant = 8.314 J mol-1 K-1
According to ideal gas law volume of a gas, V = nRT/p
Accordingly, volume of CO gas, V = 0.931 x 8.314 x 312 / 116.79 = 20.67 L
Volume (in liters) of carbonmonoxide gas required for synthesizing 29.8 g of methanol is 20.67 L.
Question 8: How many liters of oxygen (at STP) are required to form 10.5 g of H2O ? Show your work (EX 5.13)
Answer:
2 H2(g) + O2(g) <=====> 2 H2O(g)
Mass of water to be formed = 10.5 g
Moles of water to be formed = 10.5 / 18 = 0.583 moles
From balanced equation we know for 2 moles of H2O we need 1 mole of oxygen
So for 0.583 moles of H2O we need = 0.583 / 2 = 0.2915 moles of oxygen required.
From Avagadro law we know that 1 mole of a gas occupies 22.4 L of volume at STP
So 0.2915 moles will be = 0.2915 x 22.4 = 6.529 L
6.529 liters of oxygen (at STP) are required to form 10.5 g of H2O
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