Chemistry For Engineering - Class 1: 1. Gases

Chemistry for Engineering – Class 1
1. Gases:
- Pressure: is the force that results from the collisions of gas particles divided by
the area of the surface with which they collide.
- Pressure units: The SI unit of

pressure is the pascal (Pa), defined as 1N per square meter 1 Pa = 1 N/m 2.
- Other units:
o 1 mmHg = 1 torr
o 1 atm = 760 mmHg
o 1 atm = 101325 Pa = 1.01325 x 105 Pa
o 1 atm = 29.92 in Hg
o 1 atm = 14.7 psi
- The Manometer (áp kế): measures the

pressure exerted by a sample of gas.
- The ideal Gas Law:
o PV = nRT with:
 Pressure (P) in atm
 Volme (V) in L
 Moles (n) in mol
 Temperature (T) in K
 R = 0.082 atm.L/mol.K or 8.314 J/mol.K or 1.987 cal/mol.K
- Boyle’s Law:
So: P1V1 = P2V2
- Example: A
woman has an initial lung volume of 2.75 L, which is filled with air at an atmostpheric pressure of
1.02 atm. If she increases her lung volume to 3.25 L without inhaling any additional air, what is the
pressure in her lung ?
- Charles’s Law:
So: V1/T1 =
V2/T2
- Example: A sample of gas has a volume of 2.80 L at an unknown temperature.

When the sample is submerged in ice water at T = 0 oC, its volume decreases to 2.57 L. What was its
initial temperature (in K and in oC) ?
- Avogadro’s Law:
So: V1/n1 = V2/n2
- Example: A male athlete in a kinestology research study has a lung volume of
6.15 L during a deep inhalation. At this volume, his lungs contain 0.254 mol of air. During
exhalation, his lung volume decreases to 2.55 L. How many moles of gas did the athlete exhale ?
Assume constant temperature and pressure.
- Example:
o Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37
atm and a temperature of 315K.
o Calculate the number of moles of gas in a 3.24 L basketball inflated to a total
pressure of 24.3 psi at 25oC. (Note: The total pressure is not the same as the pressure read on a
pressure gauge such as the type used for checking a car or bicycle tire. That pressure, called the gauge
pressure, is the difference between the total pressure and atmostpheric pressure. In this case, if
atmostpheric pressure is 14.7 psi, the gauge pressure would be 9.6 psi. However, for calculation
involving the ideal gas law, you must use the total pressure of 24.3 psi).
- Standard condition: the volume occupied by one mole of a substance is its
molar volume, standard temperature (T = 0 oC or 273K) and pressure (p = 1.00 atm)
abbreviated as STP. So the molar volume of an ideal gas at STP is:
- Density of a Gas:
- Exampl
e:
o Calculate the density
of nitrogen gas at 125oC and a pressure
of 755 mmHg.
o A sample of gas has a
mass of 0.311 g. Its volume is 0.225 L
at a temperature of 55oC and a pressure
of 886 mmHg. Find its molar mass.
- Mixture of Gases and
Partial Pressure: The pressure due to
any individual component in a gas
mixture is its partial pressure (Pn)
- Example:
o A 1.00 L mixture of helium, neon and argon has a total pressure of 662 mmHg at
298K. If the partial pressure of helium is 341 mmHg and the partial pressure of neon is 112 mmHg,
what mass of argon is present in the mixture ?
o A 12.5 L scuba-diving tank contains a helium – oxygen (heliox) mixture made of
24.2g of He and 4.32 g of O 2 at 298K. Calculate the mole fraction and partial pressure of each
component in the mixture and the total pressure.
o Methanol (CH3OH) can be synthesized by the reaction: CO (g) + 2H2(g) ->
CH3OH(g). What volume (in liters) of hydrogen gas, at a temperature of 355K and a pressure of 738
mmHg to synthesize 35.7 g methanol ?
2. Solutions:
- A solution is a homogeneous mixture of two or more substances or
components. The majority component in a solution is typically called the solvent, and the
minority component is called the solute.
- The solubility of a substance is the amount of the substance that will
dissolve in a given amount of solvent.
- Intermolecular forces:
- Aqueous Solutions and Heats of Hydration:
- Solution Equilibrium and Factors affecting solubility:
o A solution in which the dissolved solute is in dynamic equilibrium with the solid
(undissolved) solute is a saturated solution. If you add more solute to a saturated solution, it will
not dissolve.
o A solution containing less than the equilibrium amount of solute is an unsaturated
solution. If you add additional solute to an unsaturated solution, it will dissolve.
o Temperature:
o
Pressure: The higher the pressue of a gas above the liquid the more soluble the gas is in the
liquid.
o Henry’s Law:
o Example: What pressure of carbon dioxide is required to keep the carbon dioxide
concentration in a bottle of club soda at 0.12 M at 25 oC ?
- Solution Concentration:
- Example: What
volume (in mL) of a soft drink that is 10.5% sucrose (C 12H22O11) by mass contains 78.5 g of sucrose ?
(The density of the solution is 1.04 g/L)
- Solution: CM = 10 C d/M
- Raoult’s Law:
o Example: Calculate the vapor pressure at 25 oC of a solution containing 99.5 g sucrose

(C12H22O11) and 300.0 mL water. The vapor pressure of pure water at 25 oC is 23.8 torr.
Assume the density of water is 1.00 g/mL.
o Freezing Point and Boiling Point:
o
Example:
 Calculate the freezing point of 1.7 m aqueous ethylene glycol solution.
 What mass of ethylene glycol (C 2H6O2) in grams, must be added to 1.0 kg of
water to produce a solution that boils at 105 oC ?
- Osmosis:

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Chemistry for Engineering – Class 1

- Pressure units: The SI unit of

- The Manometer (áp kế): measures the

- Example: A sample of gas has a volume of 2.80 L at an unknown temperature.

o Example: Calculate the vapor pressure at 25 oC of a solution containing 99.5 g sucrose

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