Solving emistry problems At a festival, spherical balloons with a radius of 330.

Question

Solving emistry problems At a festival, spherical balloons with a radius of 330. cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25 °C Ho) fuel are available to be burned to heat the air. calulate the maximum to 100 "C to make the balloons float 1.00 kg of butane (C and r number of balloons that can be inflated with hot air Here are some data you may find useful Specific heat capacity of air: 1.009 8 °C Density of air at 100 °C 0.946 K8 Density of butane at 100 °C: 1.898 ke Formation enthalpy of butane at 25 °C:-125.7 Any other data you need should be taken from the ALEKS Data resource Type here to search

Explanation / Answer

Radius of ballon= 330cm, 100cm= 1m, radius of ballon= 330/100=3.3 m

Volume of ballon = (4/3)*PI*r3= (4/3)*(22/7)*(3.3)3= 150.5931m3

Density of air= 0.946 kg/m3, mass of air = 150.951*0.946 kg=143 kg

This is mass of single ballon, let n= no of ballons,

Total mass of ballons= 143*n= 143n kg

The heat to be supplied= mass of air* specific heat of air* change in temperature

( specific heat of air is assumed to be independent of temperature).

=143n*1.009 KJ/kg.deg.c(100-25) =10821.53n KJ

Heat of combustion of butane is C4H10(g)+ 6.5O2 ------à4CO2(g)+ 5H2O(g)

Standard enthalpy of formation (KJ/mole): C4H10 (g)= -125.7, O2=0, CO2=-393.5 and H2O (g)= -285.8 ( formation of water vapor gives more heat of combustion compared to heat of formation of liquid water which is at -241.8 Kj/mole) and hence no of balloons will be maximum)

Standard enthalpy change= sum of standard enthalpy change of products- sum of enthalpy change of reactants= 4* standard enthalpy of formation of CO2+5* standard enthalpy of formation of H2O-{ 1* standard heat of formation of C4H10+6.5* standard heat of formation of Oxygen)

=4*(-393.5)+5*(-285.8)- { 1*(-125.7)+6.5*0} ( here 4,5, 1 and 6.5 are coefficients of CO2, H2O, C4H10 and O2 in the reaction)

=-2877.3 KJ

This heat has to be supplied from combustion of butane. Butane heat of combustion = 2877.3 Kj/mole ( data obtained)

Moles in 1 kg of butane ( molar mass of butane= 58 g/mole)= 1 Kg*1000gm/1kg/ 58 = 17.24 moles

Hence heat released from combustion of butane = moles of butane* heat of combustion of butane= *17.24 Kj=2877.3* 17.24 Kj

Hence 10821.53*n=2877.3*17.24

Hence n= 4.58 or 5

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