Question

1. What would be the new pressure (in atm.) if a gas sample at 75°C and 2.0 atm. was heated to 100.0°C at constant volume. 2. How many grams of O2 are contained in a 4.0 L sample at 2.0 atm. and 25°C? 3. What is the total pressure of a mixture of N2 and O2 if the partial pressures are 350 mmHg and 475 mmHg, respectively? What is the % of N2 and O2 in the mixture?

Answer 1

1. Or Po from Gay-Lussac's law al coustaul volume for fixed no. of moles l = constant where P is pressure and - T ů temperature (ink) Нели Р, = 2 от Ti = 757 273 = 348K P₂ = P, Tz T = 100 +273 = 373 K Pz = ? - 2 atm x 373K 12.14 atm 348K From Ideal gas equation; PV=nRT P= 2 atm, V-41, T=25C = 25 +273 = 298 K R= 0.082 1 datos/malk n = no of moles =? n= Pu - 2 atm x 41 -- 10.327 moles 0:0821 Harm x 298K malk from Dalton's law of partial pressure, Puet = P + P2 P = partial pressure of gas1 = 350 mm Hg P2 = partial pressure of gas 2 = 475 mm ty Hence, Puet = 350 +475° = 825 mm Hg Le gas in mixture - X100 % of P % H₂ = 350x100 = (42.42% % O₂ = 475 From Boyle's law for a fixed amount of gas at constant temperature, the volume occupted by the gas is inversely proportional to the pressure applied on gas or pressure of the gas ie, pra constant or Piri = P2 V2 200mL X 745 mmily = P2 X 350 mL = 425.7 min Hg Charle's Low: V = V2 => T2 = T1 V2 = (507273) 500ml 250mL 12 = 646K = (646–273)°C = 1373°C ). partial pressure Total presure ! 75 = 57.58% 825X100 V