Initial Pressure (P) = 12 atm Initial volume (V₁) = 234 Initial temperenture (T₁) = 2500.0 306K Use the combined gas law to solve the following problems: 8) If I initially have a gas at a pressure if 12 atm, a volume of 23 liters, and a temperature of 200.0 K, and then I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of the gas? V₂ = Pivita V.P, MI atm x 234X3000K Pala Fa T₁. Pa 30. L (14 atm x 200.ok). T₁ 29/57142857 30.4 T. Pa 9) A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas? = 172 initial pressure (P) = 2.3 atm Final Pressure (P₂) - 1.5 atm Final temperature (12) = 350 K initial temperature (T) 299k 30.5128205 L Final Pressure (Pa): 14atm Final volume (va) Initial temperature (1₂) P.V. - Pa Va Ta 31 L √₂-1/² x 18 ₂² Ideal Gas Law 1.Satm) 10) Calculate the pressure, in atmospheres, exerted by each of the following: a. 250 L of gas containing 1.35 moles at 320 K. PynrT mole n = 1.35 mole Temperature (1): 32ok P = nr I volume v= 250L Pressure (1): .14 atm P₁V₁12-23 atm) (172) X(350K) рахт, (299k) b. 4.75 L of gas containing 0.86 moles at 300, K. volume(v): 4.75L mole:0.86 malc Temperature: 300k Pressure: 4.5 atm V=4.75L 39.4 L P= nrT Va=312 R=0.0821 fatm mo! P= 4.459 atm T= 300k n = 0,86 mol R=0.082 La+mm (4.5 atm) 11) Calculate the volume, in liters, occupied by each of the following: a. 2.00 moles of H₂ at 300. K and 1.25 atm. 4 2.00 male X 0.0821 Latm mol K x 300k 1.25atm b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C PV=nRT, V = nRÌ_n=0.425 mole atm = 1.35 mol x 0.0821 atm mol K²X 320k 250L P=0.14a+m) = 0.86m 1X0.0821 Latm mol¹k X300 k 4.75 L RT PV=nRT V=nRT P = 39.408 39.42 6.724 atm 12) Determine the number of moles contained in each of the following gas systems: T-376-37+273 k= 310k 14.9 L 0.425 mole (NH3) X 0.0821 Latm mol¹k'x (273+37) k. 7 14.940 15884 (14.92) a. 1.25 L of O₂ at 1.06 atm and 250. K PV=ART (1.06 atm x 1,25 L) n = PV 0.0821 Latm mot klx 250 K=1 .0646 moles -0.06455542.0646 moles b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C V0.80L 030 moles, 25 atm x 0.80L) P=1₁25atm に T= 300.K = 2,00lmole R=0.082(a+mk¹ mol's P=106 atm T=250.K V=1.25L R=0.0821 L. atm molk P=0,925 atm T=27°C= 21+ 273.15=300.15 0.08211atm. mol" K'x 300.15K = 0.030029646030 Mole PV = nRT n = PV RT
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
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