\f2 V= 50dm 3 - T= Book h = M 0.05 m 3 02 - 1 kq 32 1) 02 n= 1006 = 31.25 makes He -> 1kg CO - o. 5kg He ne 1009 - 250 moles h = 299. 1 moles CO na 500 = 17.85 moles 28 2) total pressure: 3) partial pressures: PV = hRT P02 = 14 920 304 x 31.25 V= Sodm' =0.05m 299.4 ART . 299.1x 8. 314 + 300 PHe = 14 920 304 x 250 0. 05 299 . 4 PT = 14 920 309 W/m 2 PCo = 14 920 304 X 17.85 299. 43 4 litres per 1 minute; exothermic reaction 1 - molar flow rate V = 4 dm 3 = 0.004 m 3 AHcomb = - 802 kj/mol ( molar enthalpy of combustion) p = 10 N /m2 - atmospheric pressure CHy + 202 CO2 + H20 T = 25 C = 298 K PVSORT V = o. soum3 / min - V= 0.604 60 n = PV 1= 103 x0. 0000667 1 = 6.67 8. 314 x298 2427.57 n = 0. 00 26 9215 1 = 2. 692 * 10 -3 moles / molar flow rate powerrelease rate = molar flow rate x molar combustion enthalpy for CHY power = n x AHcomb [CH4]-+802 (exothermic) Power :5 P= 24 16/in2 = 24 psi - 165 474 1/m 2 winter - 5 % = 268k -> T1 summer 35 C = 308 K - T2 ASSUMPTION - volume stays constant: V1= V2 number of moles stays constant PAV1 = NAR T1 2. -> PAV ARTA n = const - = P2V2 = n2RT2 PEV MARTZ P4 T => P2 = PIX T2 P2 T2 P2 =\fpayload at 30 oooft? first: 1) What is the temperature at 30 ooo ft? 2) What is the pressure at 30 cooft? When the altitude increases by 100oft, the temperature decreases by 2"c, starting from sea level temperature . sea level - T = 25 c = 298 k 1000 ft / 2 " c 1000 : 30 000 = 2: X 30 00 0 sty x 1 1000 * = 30 000 * 2 1000 x X X T2 = 25 C - 60"C= - 35"C T2 = 238k 2) pressure decreases as altitude increases P = 4. 3psi= 30 061 Pa (30 00oft) PAYLOAD : PV= hRT (30 00oft) hel RT 30 061 X ( 13-04 => man . H 8.314 x 238 m = 1717.3x 2 n = 1717-3 moles H(He ) = 29/molASSUMPTIONS: Twater = 12ic AT = 88 C= 88K Tvapour = 100's delivery rate - 1ml /cm3 - 1ml of spray falls on lom? surface How much falls on Im 2? 1me Acm 21 1ml : x = lem2: 1m 2 X 1 m 2 * = 1me x 10 000 car2 AmL / cm 2 X = 10 000me = 1OL volume : V= 101 = 0.0 /m3 density: 8= 1000 kg/m3 m = J . V = loookg/ m3 x o. olm3 Cpwater = 4.18 5/gic Cp vapourazation at 12 c = 2 260 9/gic AHA = M CPAT = 10 x 4.18 x 88 = 3698k] D H2 = m x Cpvapourazation= * 2260 = 22 600 ky OH = AHA+ AH2 =m=0. 32129 C6H1206 Hr ( (6 H12 06) = 180. 156g/mal He(heat capacity)- 641 3/K m =o. 32/29 T1= 273. 160 k T2 = 280. 953K M -9. 3212 -0.00478 motes 180. 156 of glucose AT= 7.793k STIRRER C6H1206(5) + 60219) - 6 CO2/9)+ 6#2014) THERMOMETER COMBUSTION - EXOTHERMIC 2 = heat capacity x AT (enthalpy change is negative) 19 = 641 x 4.493 9 = 4 9957 = 4. 995 Ly ORIGEN the internal energy of combustion VESSEL II The enthalpy of combustion: WHPLE WATER A Hcomb = AHcomb 4.995 Ly 0.00148 mot & H comb = 2 806, 18 KM/ mal the enthalpy of combustion III) The enthalpy of formation exothermic reaction - Aftcomb=-2806. (8ky/ mol Atcomb = En sif products - En AHfreactants (1 - number of moles from the reaction ) A Hcomb = ( notts [ co 2 ] tho Hf [ H20] ) - ( notf [glucose] thoHy [o2] ) OHf [ CO2 ] = - 393 ky / mol AHf [ H20] = - 285 ky/mol OHS [0 2] = 0 ( in a free state ) c] the molar enthalpy of formation of glucoseBSc (HONS) FORENSIC SCIENCE Core and Materials Science The "Four candles" COURSE WORK THE FOUR CANDLES https://www.youtube.com/watch?v=OCbvCRkl_4U [1].. 48g of Oz are contained in a vessel of 15 dm3 capacity at 320K. Assuming ideality calculate:- (1) the number of moles of gas (2) the pressure of the gas (3) the density of the gas\f20 70% of chemical energy into mechanical energy 15% of mech energy for fan, water pump and alternator 85% remains it loses in the transmission ( it stays with 70% Therefore: 0. 4165 * 100% -> car efficiency\f22 T1= 293 k AT = 629.7- 293= 336.7K T2 ( boiling point) = 629. 7 kr.aim. 200,6 Cp ( Hg) = 27, 4 7 / kmol 10 kq of distillate per minute latent heat of vaporsation = 59. 27 KJ / mot -> Hrap H ( Hg) = 200. 6 9/mol m = 10 kg = 10 0oog n= 10 000 H 200. 6 power = n * ATX Cp + Huapx n power Power Power =