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Acid Base Chemistry | Gas Laws |

Memorization Test Material |

This law states that " The volume of a given amount of gas varies inversely with the pressure if the temperature is kept constant." In other word, Boyle's Law says the volume and pressure are inversely proportional at constant temperature and number of molecules. As one becomes larger the other become smaller.

Thus if p1 and V1 represent the original pressure and volume and p2 and V2 the new pressure and volume, since p x V = constant then, p1 x V1= p2 x V2 (if the temperature and amount of gas does not change.)

Example problems using Boyle's Law:

1. if 1000 mL of nitrogen aare collected at a pressure of 810 torr, what will be the volume occupied by the gas at 900 torr?

Solution: Good problem solving technique says that after reading the problem the best thing to do is set up a "Scoreboard". Therefore:

p1 = 810 torr V1 = 1000 mL

p2 = 900 torr V2 = ?

New Volume = Old Volume x Pressure Ratio

V2 = 1000 mL x 810 torr/900 torr

V2 = 900 mL

Charles found that for each degree rise in temperature the volume of all gases (even for a mixture of different gases) increase 1/273 of the volume which it occupied at 0°C. For each degree decrease in temperature the volume of all gases decrease 1/273 of the volume which it occupied at 0°C.

If a gas at 0°C is heated to 200°C, It expands 1/273 of its volume for each degree Celcius it is heated until its new volume is 200/273 greater than its volume at 0°C. By the same token, if a gas is cooled to -100°C it contracts so that its new volume is 100/273 less than its volume at 0oC.

No gas has ever been cooled to -273ºC.

All gases liquify or solidify at a temperature higher than -273ºC.

The temperature -273ºC is called absolute zero.

(Actually, -273.16ºC is believed to be the lowest possible temperature.) It is written 0 K and not 0ºK (remember K = ºC + 273).

This means that V = T x constant or V/T = k. Where T is the absolute temperature or Kelvin temperature. Using the same reasoning as we did with Boyle's Law , we can show that if a sample of gas of volume V1 and absolute temperature T1 is heated to absolute temperature T2, it expands to a new volume V2, such that : V1/T1 = V2/T2 or to simplify, V2 = V1 x T2/T1. To further simplify, as we did with Boyle's Law, we have:

New Volume = Old Volume x Temperature Ratio.

Additional rearrangement would similarly give:

New Temperature = Old temperature x Volume Ratio.

Example Problems Using Charles' Law

A sample of gas occupies 200mL at 10oC and 750mm of pressure. What is its volume at 20ºC and 750torr?

Solution: V1 = 200mL P1 = 750mm = 750 torr T1 = 10o + 273 = 283K

V2 = ? P2 = 750 torr T2 = 20o + 273 = 293K

New Volume = Old Volume x Absolute Temperature Ratio

V2 = 200ml x 293K/283K = 207mL

STANDARD TEMPERATURE AND PRESSURE

S. T. P.

Because the pressure and temperature directly affect the amount of gas present in any measured volume, we find it necessary to specify all three variables (p, V, and T), when referring to a certain amount of gas. Often, you will find, only the volume of a gas specified (example: 3.51mL of nitrogen). In such cases it is understood that the pressure of the gas is at 760torr or 1atm and the temperature is 0oC or 273K. These are the values which are accepted as STANDARD CONDITIONS (memorize) of pressure and temperature. Thes values are used throughout the scientific community of the world. The volume of a gas at these standard conditions is commonly stated to be the volume at STP (Standard temperature and pressure).

THE GENERAL (combined) GAS LAW

The volume of a confined gas may be changed either by changing the pressure or the temperature. As we have seen, the first conditioninvolves the application of Boyle's Law and the second involves Charles' Law. However, a change of volume, pressure, and temperature can (and does) occur at the same time. Therefore, we find it convenient to combine the two equations associated with Boyle and Charles into a single equation which we will call the GENERAL GAS LAW.

Since: Boyle Law states p1 x V1= p2 x V2

Charles' Law states V1/T1 = V2/T2

Therefore Combined V1p1/T1 = V2p2/T2

Where: V1 = initial volume V2 = final volume

p1 = initial pressure p2 = final pressure

T1 = initial absolute temperature T2 = final absolute temperature

New Volume = Old Volume x Pressure Ratio x Temperature Ratio

New Pressure = Old Pressure x Volume Ratio x Temperature Ratio

New Temperature = Old Temperature x Volume Ratio x Pressure Ratio

Example Problems Using The General Gas Law:

Suppose that a sample of gas is found to occupy 500mL under the laboratory conditions of 27ºC and 740torr. Correct the volume to STP (or 0ºC and 760torr).

Solution: First set up your "scoreboard" and convert all temperatures to an absolute scale, such as Kelvin.

V1 = 500mL p1 = 740torr T1 = 27ºC + 273 = 300K

V2 = ? p2 = 760torr T2 = 0ºC + 273 = 273K

Since there is a decrease in temperature there must also be a proportional decrease in volume, thus we use a temperature ratio of 273K/300K. We also see that there is an increase in pressure which will produce a further decrease in volume, therefore we use the ratio of 740torr/760torr. Now we can multiply the original volume by the temperature ratio and the pressure ratio to obtain the new volume at STP.

New Volume = Old Volume x Pressure Ratio x Temperature Ratio

New Volume = 500mL x 740torr/760torr x 273K/300K = 443mL

DALTON'S LAW OF PARTIAL PRESSURES

In a mixture of different gases each gas exerts part of the pressure. This is the same pressure that it would exert if it alone occupied the volume containing the mixture of gases. In 1803 John Dalton formulated the Law:

"THE TOTAL PRESSURE IN A MIXTURE OF GASES IS THE SUM OF THE INDIVIDUAL PRESSURES".

Stated mathematically PT = p_{1} + p_{2} + . . . . + p_{n} where the PT is the total pressure and the small p's refer to the part of the pressure exerted by each individual gas in the mixture.

Dalton said that the same thing would happen if equal volume of gases which were under different pressures were put into a single container.

760torr + 80cm Hg + 750mm Hg = 906cm Hg (or 9060torr)

Example Problems Using Dalton's Law:

A sample of air held in a graduated cylinder over water, has a volume of 88.3mL aat 18.5ºC and a pressure of 741torr. What would be the volume of the air if it were dry and at the same temperature and pressure?

Solution:

The trapped air is saturated with water vapor. The total pressure of 741 is the sum of the pressure of the air plus the pressure of the water vapor. At 18.5ºC the vapor pressure of water is 16.0torr (from the Handbook of Chemistry and Physics or any vapor pressure table).

(Dalton's Law) Ptotal = pdry air + pwater vapor

OR

pdry air = Ptotal - pwater vapor

pdry air = 741torr - 16torr = 725torr

P1 = 725torr V1 = 88.3mL T1 = 18.5ºC

p2 = 741torr V2 = ? T2 = 18.5oC

Note that in the "changed" condition the dry air exerts the entire pressure of 741torr. (Think of it as reaching in and removing the water molecules.) Consequently, the volume of the dry air must be smaller and is reduced by the ratio of the pressure change (725torr/741torr). Thhere is no change in the temperature, so the temperature ratio can be ignored. We have only to consider Boyle's Law:

New Volume = 88.3mL x 725torr/741torr = 86.4mL

UNIVERSAL GAS LAW

(Ideal gas law)

We have just seen that the volume of a gas can vary by a change in pressure, temperature, and/or number of molecules of the gas. The volume can cahnge in proortion to 1/p, T, and/or n. From this we can derive that volume is proportional to 1/p x T x n.

[V &Mac176; 1/p]T,n

[V &Mac176;T]p,n

[ V &Mac176; n ]p,T

_________

[V &Mac176; 1/p.T.n] (notice nothing has to be kept constant)

OR: V =T.n/p x constant

Since in this case the constant is not dependent on anything else being kept constant it is a TRUE CONSTANT and is given its own symbol "R". Thus, the Universal Gas Law Equation would be written V =T.n.R/p Or by multiplying both sides by p: pV=nRT (memorize this)

Where: p = pressure in atmospheres

V = volume in liters

n = number of moles

T = absolute temperature on the Kelvin scale

R is the UNIVERSAL GAS LAWS CONSTANT

The numerical value, along with its units, for R can be calculated by substituting the values of p, V, n, and T, at standard conditions into the equation just derived.

p = 760torr or 1 atm n = 1 mole T = 273K

V = 22.4L (From one mole of any gas at STP occupies 22.4 L)

From pV = nRT , by dividing both sides by nT we have R = pV/nT. Then by substituting we get:

R = 1 atm x 22.4 L Or R = 0.08205 L atm = UNIVERSAL GAS CONSTANT

1 mole x 273 K mol K

MEMORIZE THE CONSTANT INCLUDING VALUE AND UNITS!!!!!!!!!

Note: n = number of moles = actual mass of substance

gram formula mass

Before we continue, let us see if we can go back to the other gas laws from pV = nRT. Remember that in the other gas laws we always had a befor aand after set of data. So we can set up a before and after with pV = nRT, pV = nRT (case one) .

pV = nRT (case two)

Since R1 = R2 we have: pV = nT (case one) .

pV = nT (case two)

Now if n1 = n2 we have the general gas law: p1V1 = T1

p2V2 T2

Similarly if n and T are both constant and cancel we have Boyle's law equation, or if n and p cancel we will have Charles' law equation. In some of the problems which follow you will have to do just this; so now you have been forewarned. pV = nRT, the UNIVERSAL GAS LAW , can also be used ie, and with, ratio and proportion. (Schedule Dumas Experiment)

Sample Problem Using The Universal Gas Law:

Under the conditions of 27oC and 720torr what would be the volume occupied by 7.31g of carbon dioxide?

Solution: A study of the problem will show that we are given mass, pressure, and temperature of a gas and from the formula we can calculate its molecular weight (molar mass). The volume is the unknown variable. Therefore we can use the universal gas law equation, pV = nRT, to solve the problem. Solving the equation for the unknown variable, V, we get V = nRT/p. To use the value 0.08205 L atm/mol K for R, we must measure n in moles, T in Kelvin, and p in atmospheres. The value of V will be determined in liters.

p = 720torr or 720/760 atm = 0.947 atm m = 7.35 g gramformula mass = 44.0 g/mol

T = 27oC + 273 = 300K n = 7.31g/44.0g/mol = 0.166mol

We can now solve for the volume V in liters:

V = 0.166mol x 0.08205 L atm/mol K x 300K = 4.31 L

0.947 atm

Or, to save time aand accuracy we can do chain arithmetic with the calculator:

V = nRT/p = (7.31g/44.0g/mol) x 0.08205L atm/mol K x 300K

(720torr/760torr/atm)

This simplifies into V = 7.31g mol x 0.08205L atm x 300K x 760torr = 4.32L

44.0g x mol K x 720torr atm

Using gas laws to solve stoichiometry problems. Lets use the energy problems as an example:

example: Given 50.0 grams of sodium bicarbonate and 33.3 kJ of energy, how many mL of carbon dioxide can be collected over condensed water assuming we cool the final temperature to 4.0 ºC and the pressure is 677.7 mmHg. Because energy is now consided a reactant and you have "two" reactants this is a limiting reactant problem.

reaction: 123 kJ + 2NaHCO_{3} --------------> Na_{2}CO_{3} + H_{2}O + CO_{2}

since we determine for 50.0 g NaHCO_{3} that 36.6 kJ of energy was necessary, with only 33.3kJ energy is now the limiting reactant therefore we start the problem with energy and solve to the moles of the product we are interested in and plug this into the ideal gas law equation and do not forget to subtract the water vapor pressure from the atmospheric pressure.

33.3kJ x 1mole CO_{2}/123kJ =0.27 mole CO_{2} Now plug the moles into ideal gas law equation pv=nrT

we are solving for v therefore v=nrT/p water vapor at 4.0 ºC is 6.1 torr

v= (0.27moles *(62.4 L* torr/K)* (277.0K))/(677.7 torr -6.1 torr)= 6.95 L of CO_{2} or 6.95 x103 mL