Physical Chemistry CHEM 315 EXAM 1 &2 2 November 201 Max: 125 Points Time: 2 HOU
ID: 553270 • Letter: P
Question
Physical Chemistry CHEM 315 EXAM 1 &2 2 November 201 Max: 125 Points Time: 2 HOURS Instructions: This test is worth more than 125 points. You have the option to "pick and choose" the questions that you'd wish to answer; your maximum score will however be 125 ONLY! There is no partial credit on any question. Please report your answer only in the space provided below each question Answers with no units or wrong units shall fetch you NO credit! All the questions are based upon the concepts and material discussed in the class Each question is worth 6 points, unless mentioned otherwise. 1. An inflated balloon has a volume of 6.0 dm2 at sea level(1.0 atm)jand is allowed to ascend until the pressure is 0.45 atm During ascent, the temperature of the gas falls from 22 °C to -21 C) What is the volume of the balloon at its final aititude? 2. In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen irn the presence of a suitable catalyst to form nitric oxide and water vapour: 4 NH3 (g) + 5 O2 (g)-- with 1.00 mol of oxygen at 850 °C and 5.00 atm in this reaction? -> 4 NO (a) + 6 HsO (g). What volume of ammonia would react vessel of volume 22.4 dm2 contains 20 moles of Ha gas and 1.0 mole of Na gas at temperature of o oC. Calculate theirp also the total pressure in that vessel 4. If a gas in an aerosol container exerts a pressure of 125 kPa at a temperature of 18 OC, what would the pressure inside this container be, had it be thrown into a furnace at a temperature of 700 0C? 5. Calculate the pressure exerted 1.0 mole of ethane behaving like (a) a perfect gas (b) as a real gas under the conditions of 0 OC and 22.414 dm 6. Calculate the rms velocity of O2 molecules at 30 °C.Explanation / Answer
4) we know that,
P1/T1 = P2/T2
T1= 18 + 273 = 291 K , T2 = 700 + 273 = 973 K
125 kPa/ 273 K = P2/973 K
P2 = (125 kPa * 973 K)/ 273 K = 445.5 kPa
5) a) for Perfect gas : PV= nRT
given n = 1 mol, T = 273 K , V= 22.414 dm3 = 22414 cm3 , R = 82.057 cm3 atm / K . mol
P = (1 mol * 82.056 cm3 atm / K . mol * 273 K ) / 22414 cm3 = 0.999 atm = 1 atm (apprx)
b) for real gas :
(P + an2/V2) (V-nb) = nRT
where a = 5.51 L2atm/mol2 , b = 0.0651 L/mol , V = 22.414 dm3 = 22.414 L, n = 1 mol , T = 273 K, R = 0.082 Latm /K mol
an2/V2 = 5.51 L2atm/mol2 * 1 mol2 / 502.3 L2 = 0.01096 atm
V-nb = 22.414 L - (1 mol * 0.0651 L/mol )= 22.3489 L
(P +0.01096 atm ) * (22.3489 L) = (1 mol * 0.082L atm / K . mol * 273 K )
(P +0.01096 atm ) = (22.386 L .atm) / (22.3489 L) = 1.00166 atm
P = 1.00166 atm - 0.01096 atm = 0.9907 atm
6) RMS velocity calculation for Oxygen molecule
Vrms = (3RT/M)1/2
M = Mass of the molecule in kg = 32 g/mol = 32 x 10-3 kg/mol = 3.2 x 10-2 kg/mol
R = 8.3145 (kg.m2/sec2 / K . mol)
T = 30 + 273 = 303 K
Vrms =[ 3 * 8.3145 (kg.m2/sec2 / K . mol) * 303 K / 3.2 x 10-2 kg/mol]1/2
Vrms = 485 m/sec