Gas Laws & Conversions[1]

PV=nRT System 1 :

System 2 :

Pressure: atm, Volume: Litre, R : 0.0821atm*l/mol*K, Temperature : kelvin

Pressure: Pa(pascals or N/m2) , Volume: metre cube (m3) , R : 8.314 Joule/mol*K, Temperature : kelvin

Unit

Pressure TIP

Unit

atm bar N/m2; Pa Kn/m2;Kpa

Use system 1 convert to atm Use system 2 convert to system 2

Litres Use system 1 m3 Use system 2 ml;mililitre convert to Litres cc;cm3 convert to litres

Pressure conversions Units of Pressure 1 pascal (Pa); 1N/m2 1 atmosphere (atm) 1 atmosphere (atm) 1 bar 1 torr

Volume TIP

Moles( n) TIP

Unit -

ok ok ok ok

Unit atm*l/mol*k cal/mol*k J/mol*k

Gas Constant conversions

1 N*m-2 = 1 kg*m-1*s-2 1.01325*10^5 Pa(N/m2) 760 torr 105 Pa 1MM OF BAROMETER Height

Gas Constant ('R) Value TIP 0.082 2 8.324

Use system 1 conver to joules(*4.128) use system 2

Unit

Kelvin(K) Celsius

=> 0.082 atm*L = 8.314 J How????

ok convert to kelvin

Volume conversions

1m3 1L 1ml

0.082 ATM*L/mol*K=8.314 J/mol*K

Temperature(T) TIP

1000L 1000ml 1 cm3

Gas Laws Gases behave differently from the other two commonly studied states of matter, solids and liquids, so we have different methods for treating and understanding how gases behave under certain conditions. Gases, unlike solids and liquids, have neither fixed volume nor shape. They are molded entirely by the container in which they are held. We have three variables by which we measure gases: pressure, volume, and temperature. Pressure is measured as force per area. The standard SI unit for pressure is the pascal (Pa). However, atmospheres (atm) and several other units are commonly used. The table below shows the conversions between these units. Units of Pressure 1 pascal (Pa) 1 N*m-2 = 1 kg*m-1*s-2 1 atmosphere (atm) 1.01325*105 Pa 1 atmosphere (atm) 760 torr 1 bar 105 Pa Volume is related between all gases by Avogadro's hypothesis, which states: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. From this, we derive the molar volume of a gas (volume/moles of gas). This value, at 1 atm, and 0° C is shown below. Vm = V n = 22.4 L at 0°C and 1 atm Where: Vm = molar volume, in liters, the volume that one mole of gas occupies under those conditions V=volume in liters n=moles of gas An equation that chemists call the Ideal Gas Law, shown below, relates the volume, temperature, and pressure of a gas, considering the amount of gas present. PV = nRT Where: P=pressure in atm T=temperature in Kelvins R is the molar gas constant, where R=0.082058 L atm mol-1 K-1. The Ideal Gas Law assumes several factors about the molecules of gas. The volume of the molecules is considered negligible compared to the volume of the container in which they are held. We also assume that gas molecules move randomly, and collide in completely elastic collisions. Attractive and repulsive forces between the molecules are therefore considered negligible. Example Problem: A gas exerts a pressure of 0.892 atm in a 5.00 L container at 15°C. The density of the gas is 1.22 g/L. What is the molecular mass of the gas? Answer: PV = nRT T = 273 + 15 = 228 (0.892)(5.00) = n(.0821)(288) n = 0.189 mol .189 mol 5.00L x x grams 1 mol = 1.22 g/L x = Molecular Weight = 32.3 g/mol We can also use the Ideal Gas Law to quantitatively determine how changing the pressure, temperature, volume, and number of moles of substance affects the system. Because the gas constant, R, is the same for all gases in any situation, if you solve for R in the Ideal Gas Law and then set two Gas Laws equal to one another, you have the Combined Gas Law: P1V1 n1T1 = P2V2

n2T2Where: values with a subscript of "1" refer to initial conditions values with a subscript of "2" refer to final conditions If you know the initial conditions of a system and want to determine the new pressure after you increase the volume while keeping the numbers of moles and the temperature the same, plug in all of the values you know and then simply solve for the unknown value. Example Problem: A 25.0 mL sample of gas is enclosed in a flask at 22°C. If the flask was placed in an ice bath at 0°C, what would the new gas volume be if the pressure is held constant? Answer: Because the pressure and the number of moles are held constant, we do not need to represent them in the equation because their values will cancel. So the combined gas law equation becomes: V1 T1 = V2 T2 25.0 mL 295 K = V2 273 K V2 = 23.1 mL We can apply the Ideal Gas Law to solve several problems. Thus far, we have considered only gases of one substance, pure gases. We also understand what happens when several substances are mixed in one container. According to Dalton's law of partial pressures, we know that the total pressure exerted on a container by several different gases, is equal to the sum of the pressures exerted on the container by each gas. Pt = P1 + P2 + P3 + ... Where: Pt=total pressure P1=partial pressure of gas "1" P2=partial pressure of gas "2" and so on Using the Ideal Gas Law, and comparing the pressure of one gas to the total pressure, we solve for the mole fraction. P1 Pt = n2 RT/V

nt RT/V = n1 nt = X1 Where: X1 = mole fraction of gas "1" And discover that the partial pressure of each the gas in the mixture is equal to the total pressure multiplied by the mole fraction. P1 = n1 nt Pt = X1Pt Example Problem: A 10.73 g sample of PCl5 is placed in a 4.00 L flask at 200°C.

a) What is the initial pressure of the flask before any reaction takes place? b) PCl5 dissociates according to the equation: PCl5(g) --> PCl3(g) + Cl2(g). If half of the total number of moles of PCl5(g) dissociates and the observed pressure is 1.25 atm, what is the partial pressure of Cl2(g)? Answer: a) 10.73 g PCl5 x 1 mol 208.5 g = 0.05146 mol PCl5 PV = nRT T = 273 + 200 = 473 P(4.00) = (.05146)(.0821)(473) P = 0.4996 atm .05146 mol 0 mol 0 mol Change: -.02573 mol +.02573 mol +.02573 mol Final: .02573 mol .02573 mol .02573 mol ntotal = PCl2 Ptotal

PCl2

1.25 atm = .02573 mol

.07719 mol

PCl2 = .4167 atm

b) PCl5 → PCl3 + Cl2 XCl2 = nCl2

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