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Ptotal = P 1 + P 2 + P3 ...Gas Law Formulas Dalton’s Law of Partial Pressure X 1 = n V 1 P / T 1 /n 1 V 1 1 total = P= V = P 22 V / T 21 /P 2 total Mole FractionCharles’ LawBoyle’s Law V P 11 / n/ T 1 1 = V= P 22 / n/ T (^2 2) Gay-Lussac’s LawAvogadro's Law P 1 V 1 / T 1 = P 2 V 2 / T 2 Combined Gas Law Graham's Law R = 8.3145 L kPa/mol K orR= 0.08206 L atm/mol K^ PV = nRT Ideal Gas Law R= 0.08206 L atm/mol Kmm = molar mass^ (mm) P = dRTd = density^ Gas Density/Molar Mass
M = molar mass in kg / molR = 8.3145 J/mol K^ v^ rms^ =^ √(3RT / M)^ Root Mean Square Velocity
[Pobs + a(n/V)1 atm = 760 torr = 760 mm Hg = 101.3 kPa = 14.7 psi^2 ] x (V – nb) = nRT Standard Atmospheric Pressure: van der Waals Equation
Gases Gases have four physical properties that we regularly measure:
P V T = pressure (measured in atm)= temperature (measured in K)= volume (measured in L) n Historically, at least three scientists performed experiments th = amount (measured in moles) at brought us the ideal gas
law we use today: Boyle’s Law
Boyle did an experiment where he kept the amount of gas and the temperature constant, and varied the volume and the pressure. Let’s say he did something like this (NOTE this is NOT his actual experiment!): –
P 1
P 2
Where he put the same amount of gas in each container (n was constant) and held both containers at the same temperature (T was constant). Then he measured the pressure inside each of the containers. Like any good scientist, he performed this experiment a whole bunch of times and graphed his results, which looked something like this:
As far as graphs in chemistry go, this graph sucked. It did not give a direct proportionality between the two variables P and V AND it had the possibility of asymptotes (where at a given value of P, V would approach infinity). NOT GOOD! So, he played with the graph a bit by taking the inverse of the pressure and got the following:
V 1 V 2
His new origin he called 0 degrees Kelvin, an to find the temperature in Kelvin (which you must do for all gases relationship does not work), you simply add 273. 15 to the Celsius temperature:d it was equal to exactly – otherwise the - 273.15°C. Thus, And in this case:^ TK^ = T°C^ + 273. V∝T !
V T ∝ 1
V T = constant !
V T 11 = V T 22 (this is Charles’s Law)
Avogadro’s Law Avogadro did an experiment where he kept the temperature and the pressure constant, and varied the volume and the amount of gas. Let’s say he did something like this
(NOTE – this is NOT his actual experiment!): n 1
n 2
Where he put different amounts of gas in each container but held both containers at the same pressure (P was experiment a whole bunch of times and graphed his results, which were a straight line constant) and temperature (T was constant). He performed this that began at the origin (FINALLY!). So: V∝n !
V n ∝ 1 !
V n = constant !
V n 11 = V n 22 (this is Avogadro’s Law)
V 1 V 2
Combined Gas Equation If we decided that each of these scientists had used the exact same volumes, then we could combine all three laws to give:
T^ P^11 Vn 11^ =^ TP 22 Vn^22 YOU SHOULD MEMORIZE THIS EQUATION! same variable in a problem (i.e. two pressure values given in a problem). If a variable is said to be held constant or is not mentioned, then eliminate that variable from the Use it anytime you have two of the equation altogether.
Ideal Gas Law In the variables was equal to a constant. The same remains true here: previous laws, every time we looked at one side of the equation, the half with the
PV Tn = R Where R is known as the Ideal Gas constant: ! The equation given above, known as the ideal gas law, is more commonly seen as:^ R^ =^^0.^0821 mole KL atm YOU SHOULD MEMORIZE THIS EQUATION TOO!^ PV = nRT Use it anytime you have only one of three of these variables ( as your conversion f MUST match those ofactor, so when you use this equation, the units of the variables R. P, V, n, or T). You’ll notice that the units of R act
Problems: 1. A sample of Freon volume at STP (where STP is standard temperature and pressure-12 (CF 2 Cl 2 ) occupies 25.5 L at 298 K and 153.3 kPa. Find its – 0 °C and 1 atm).
- [101.325 kPa = 1 atm]How many moles of gaseous arsine (AsH the density of gaseous arsine? 3 ) will occupy 0.0400 L at STP? What is
- When an evacuated 63.8 the bulb gains 0.103 g in mass. Is theHow many grams of phosphine (PH-mL glass bulb is filled with a gas at 22 3 ( (^) g gas N)) can form when 37.5 g of phosphorus (P 2 , Ne, or Ar? [760 mm Hg = 1 atm]°C and 747 mm Hg, 4 ( s )) and 83.0 L of hydrogen gas (H 2 ( g )) react at STP?
Problem Solution To solve this problem, first label the numbers in the problem with the correct variable (your choices: P, V, n, or T). Choose the variable by looking at the units next to the number in the problem. n = 1.50 mole V T P == ?= 3.30 L 25 °C + 273 = 298 K
! Remember, in gas problems, we MUST convert temperatures to Kelvin (T^ R^ =^^0.^0821 mole KL atm K = T°C + 273). Because there is one of each variable listed above, we need to use the ideal gas law: PV = nRT Solve for P : ! So, the answer is^ P^ = E^ nRT.^ V^ =^ (1.50 mole x 0.0821^ 3.30 L^ mole KL atm^ x 298 K) =^^11.^1 atm Worked Out Solution Example 1. Find the volume of 0.110 M hydrochloric acid necessary to react completely with Problem Solution^ 1.52 g Al(OH)^3 To solve this problem, first write out the balanced chemical equation: 3 HCl(aq) + Al(OH) 3 (s) AlCl 3 (aq) + 3 H 2 O(l) Then, evaluate your numbers given. We have 0.110 M of HCl, which gives us moles per L and can be used to find a volume, bu use the amount given of Al(OH) 3 to find the moles of HCl:t we need moles of HCl first. Therefore, we must
! Finally, use the moles of HCl and the molarity of HCl to find the volume of HCl:^1.^52 g Al(OH)^3 x^78 1 mole Al(OH). 01 g Al(OH)^33 x^ 1 mole Al(OH)3 mole HCl^3 =^ 0.0585 mole HCl !
0. 0585 mole HCl x (^) 0.110 mole HClL HCl = 0.531 L HCl or 531 mL HCl
