1.

In a constant volume calorimeter, 3.5g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0K to 298.45 K due to the combustion of the gas in kJ mol^(-1) is

Answer»

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Solution :`n=(3.5)/(28)`
`DeltaT=T_(2)-T_(1)=298.45-298=0.45`
`C_(P)=C_(V)+R=2500+8.314=2508.314 JK^(-1)`
`Q_(P)=C_(P)DeltaT=1128.74 J`
`DeltaH=(Q_(P))/(n)=(1128.74)/(3.5//28)`
implies 9030 J `mol^(-1)`=9.030 kJ `mol^(-1)` = 9 kJ `mol^(-1)`.


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