How much heat will be absorbed by $1$ mole of an ideal gas, if it is expanded reversibly from $5 \mathrm{L}$ to $25 \mathrm{L}$ at ${27}^o\mathrm{C}$? 

$(\log 3=0.5, \log 4=0.6, \log 5=0.7)$

How much heat will be absorbed by $1$ mole of an ideal gas, if it is expanded reversibly from $5 \mathrm{L}$ to $25 \mathrm{L}$ at ${27}^o\mathrm{C}$? 

$(\log 3=0.5, \log 4=0.6, \log 5=0.7)$

When $0.8 \mathrm{g}$ of glucose, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$, was burnt in a bomb calorimeter, according to the following equation, the temperature raise is found to be $2 \mathrm{K}$ at $1 \mathrm{atm}$.

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_{6\left(\mathrm{s}\right)}+6{\mathrm{O}}_{2\left(\mathrm{g}\right)}⟶6{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}$

If the heat capacity of the bomb calorimeter is $6.8 \mathrm{kJ} {\mathrm{K}}^{-1}$, the approximate enthalpy change of the reaction in $\mathrm{kJ} {\mathrm{mol}}^{-1}$ is (Molar mass of glucose $=180 \mathrm{g} {\mathrm{mol}}^{-1}$ )