1. Tardigrade
  2. Question
  3. Chemistry
  4. In a constant volume calorimeter. 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increases from 298.0 K to 98.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K -1. the numerical value for the enthalpy of combustion of the gas in kJ mol -1 is

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Q. In a constant volume calorimeter. $3.5\, g$ of a gas with molecular weight $28$ was burnt in excess oxygen at $298.0\, K$. The temperature of the calorimeter was found to increases from $298.0 \,K$ to $98.45\, K$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 \,kJ\, K ^{-1}$. the numerical value for the enthalpy of combustion of the gas in $kJ\, mol ^{-1}$ is

IIT JEEIIT JEE 2009Thermodynamics

Solution:

Temperature rise $=T_{2}-T_{1}=298.45-298=0.45 \,K$
$q=$ heat capacity $\times \Delta T=2.5 \times 0.45=1.125 \,kJ$
$\Rightarrow $ Heat produced per mole $=\frac{1.125}{3.5} \times 28=9 \,kJ$