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