Can you help me with the number 12 question? Can you explain step by step including the formula (Ideal Gas Law)? I need to plug in the fraction to give the correct answer. Initial Pressure (P₁) = 12 atmInitial volume (V₁) = 234Initial temperenture (T₁) = 2500.0Use 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, andthen I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of theV.P, MIgas?V₂= Pivitaтраatm x 23LX3000KPavaFa30. L(14 atm x 200.0k).T₁29/5714285730.49) 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 theva-PivitaT. Patemperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?volume = 172initial pressure (P) = 2.3atm Final Pressure (Pa) = 1.5 atminitial temperature (T) 299KFinal temperature (12) = 350 K30.5128205 LFinal Pressure (Pa): 14atmFinal volume (Va) - 300KInitial temperature (1₂)P.V₁ - Pa VaТа31 L√₂=RK x JBPiVita - 3 atm) (174) X(350K)Рахті(2991)(1,Satm)Pa v₂=Ideal Gas Law39.4 L10) Calculate the pressure, in atmospheres, exerted by each of the following:a. 250 L of gas containing 1.35 moles at 320 K. Pynrt R=0.0821 fatm mo'k!mole n = 1.35 mole Temperature (1): 320k P = nr Ivolume v= 250L Pressure (1):.14 atmP=nrtb. 4.75 L of gas containing 0.86 moles at 300, K.volume (v): 4.75L mole:0.86 MolecTemperature 300k Pressure:P= 4.459 atm4.5 atmV=4.75Ln=0.86mok, R=0.082 Latmma1 / 14.5 atm)T= 300k11) Calculate the volume, in liters, occupied by each of the following:a. 2.00 moles of H₂ at 300. K and 1.25 atm.42.00mde X 0.0821 Latm mol K x 300k1.25atmb. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°CPV=nRT, V = nRÌ_n=0.425 moleatm12=Va= 312P=0.14 atm= 0.86m 1X0.082 | Latm mol k X300K4.75 La= 1.35 mol x 0.082 1 atm mol K²X 320k250LRTPV=nRTn=V=MRI= 39.40839.42T-376-37+273 k= 310k14.9 L 0.425 mole (NH3) X 0.0821 Latm mol¹k'x (273+37)k.a. 1.25 L of O₂ at 1.06 atm and 250. K PV=ART(1.06 atm x 1,25 L)n = PV0.0821 Latm mot klx 250 K=1.0646 moles-0.06455542.0646 molesb. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C V0.80L030 moles, 25 atm x 0.80L)P=125atmにT= 300.K6.724 atm12) Determine the number of moles contained in each of the following gas systems:P=106 atm T=250.kV=1.2521=2,00lmoleR=0.082(a+mk mol's714.940 15884(14.92)R=0.08212. atm molkP=0,925 atmT=27°C= 21+ 273.15=300.150.08211atm. mol" K'x 300.15K = 0.030029646030 MolePV = nRTn = PVRT

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Can you help me with the number 12 question? Can you explain step by step including the formula (Ideal Gas Law)? I need to plug in the fraction to give the correct answer. Initial Pressure (P₁) = 12 atmInitial volume (V₁) = 234Initial temperenture (T₁) = 2500.0Use 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, andthen I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of theV.P, MIgas?V₂= Pivitaтраatm x 23LX3000KPavaFa30. L(14 atm x 200.0k).T₁29/5714285730.49) 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 theva-PivitaT. Patemperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?volume = 172initial pressure (P) = 2.3atm Final Pressure (Pa) = 1.5 atminitial temperature (T) 299KFinal temperature (12) = 350 K30.5128205 LFinal Pressure (Pa): 14atmFinal volume (Va) - 300KInitial temperature (1₂)P.V₁ - Pa VaТа31 L√₂=RK x JBPiVita - 3 atm) (174) X(350K)Рахті(2991)(1,Satm)Pa v₂=Ideal Gas Law39.4 L10) Calculate the pressure, in atmospheres, exerted by each of the following:a. 250 L of gas containing 1.35 moles at 320 K. Pynrt R=0.0821 fatm mo'k!mole n = 1.35 mole Temperature (1): 320k P = nr Ivolume v= 250L Pressure (1):.14 atmP=nrtb. 4.75 L of gas containing 0.86 moles at 300, K.volume (v): 4.75L mole:0.86 MolecTemperature 300k Pressure:P= 4.459 atm4.5 atmV=4.75Ln=0.86mok, R=0.082 Latmma1 / 14.5 atm)T= 300k11) Calculate the volume, in liters, occupied by each of the following:a. 2.00 moles of H₂ at 300. K and 1.25 atm.42.00mde X 0.0821 Latm mol K x 300k1.25atmb. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°CPV=nRT, V = nRÌ_n=0.425 moleatm12=Va= 312P=0.14 atm= 0.86m 1X0.082 | Latm mol k X300K4.75 La= 1.35 mol x 0.082 1 atm mol K²X 320k250LRTPV=nRTn=V=MRI= 39.40839.42T-376-37+273 k= 310k14.9 L 0.425 mole (NH3) X 0.0821 Latm mol¹k'x (273+37)k.a. 1.25 L of O₂ at 1.06 atm and 250. K PV=ART(1.06 atm x 1,25 L)n = PV0.0821 Latm mot klx 250 K=1.0646 moles-0.06455542.0646 molesb. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C V0.80L030 moles, 25 atm x 0.80L)P=125atmにT= 300.K6.724 atm12) Determine the number of moles contained in each of the following gas systems:P=106 atm T=250.kV=1.2521=2,00lmoleR=0.082(a+mk mol's714.940 15884(14.92)R=0.08212. atm molkP=0,925 atmT=27°C= 21+ 273.15=300.150.08211atm. mol&#34; K'x 300.15K = 0.030029646030 MolePV = nRTn = PVRT

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