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A comprehensive overview of the fundamental gas laws, including dalton's law of partial pressure, boyle's law, charles' law, avogadro's law, gay-lussac's law, and the combined gas law. It also introduces the ideal gas law and the van der waals equation, as well as discusses the physical properties of gases and the historical experiments that led to the development of these gas laws. Several worked-out examples and practice problems to help students understand the application of these gas law formulas in various scenarios. This resource would be particularly useful for students studying chemistry, physics, or engineering, as it covers the essential concepts and equations needed to analyze and solve problems related to the behavior of gases.
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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
[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
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
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!): –
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:
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 = constant !
V T 11 = V T 22 (this is Charles’s Law)
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)
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.
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
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