Recent content by mojo4king

closed system,frictionless piston initially filled with gas molecules A, piston mass 100g,area 10cm2,10cm from bottom,pressure outside 1atm,temp. iniside system 900oC, i have already calculated the initial number of moles using pv=nrt... Next using the collision flux multiplied by the area i...

Hello, Piston question,i need to work out the time when no molecules are left in the gaseous system.. I have worked out the relative velocity..if i multiply that by the mean free path to get the diffusion coefficient am i getting any closer to the answer? I can work out the change in time...

however i need to work out the CO2 output per unit mass and volume, the units would be kj/kg and kj/m3,correct? the CO2 energy output per mass... -802.34 kj/mol divided by 12 g/mol equals 66860 kj/kg the CO2 energy output per unit volume: 66860 x 0.645 (CH4 mass density)= 43124.7 kj/m3 Have i...

-802.34 kj/mol divided by 12 g/mol equals 66.86 kj/g CORRECTION: DIVIDED BY 16G/MOL

for eg. combustion of CH4: CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) enthalpy of combustion: (-393.51) + 2(-241.82) - (-74.81) = -802.34 kj/mol energy output per unit mass: 802.34 x 1000/16(molar mass) = -50146.25 kj/kg energy output per unit volume: -50146.25 x 0.645(mass density) = -32344.33...

1 mole of CO2= 44g 1 mole of CH4= 16g At STP the volume of an ideal gas occupies 0.0224 m3...

for eg. combustion of CH4: CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) enthalpy of combustion: (-393.51) + 2(-241.82) - (-74.81) = -802.34 kj/mol energy output per unit mass: 802.34 x 1000/16(molar mass) = -50146.25 kj/kg energy output per unit volume: -50146.25 x 0.645(mass density) =...

for eg. combustion of CH4: CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) enthalpy of combustion: (-393.51) + 2(-241.82) - (-74.81) = -802.34 kj/mol energy output per unit mass: 802.34 x 1000/16(molar mass) = -50146.25 kj/kg energy output per unit volume: -50146.25 x 0.645(mass density) = -32344.33...

Carbon dioxide output?? How would you go about calculating the carbon dioxide output per unit mass and volume for combustion products, i have already calculated the energy outputs per unit mass and volume. Many thanks.

Ive been trying to figure this one out for too long,please help :s ! Molar volume of a saturated vapour-liquid system is given by V=Vf(1-X) + VgX, where Vf=molar vol. of condensed phase Vg=molar volume of vapour phase X=quality A given closed system contains only water in VLE at 25oC...