Calculate $\Delta H$ for the following reaction at $300 \text{ K}$: $2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$ if $\Delta U$ for the reaction is $-80 \text{ kJ}$ $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$ (in $\text{ kJ}$)

What is the difference between $\Delta H$ and $\Delta U$ for the reaction given below at $298 \ K$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$

$A$ gas undergoes a change from state $A$ to state $B$. In this process,the heat absorbed and work done by the gas are $5\, J$ and $8\, J$,respectively. Now,the gas is brought back to $A$ by another process during which $3\, J$ of heat is evolved. In this reverse process of $B$ to $A$:

The mathematical equation of the first law of thermodynamics for an isochoric process is:

Calculate the change in internal energy of the system if $20 \ kJ$ work is done on the system and it releases $10 \ kJ$ heat in a particular reaction. (in $kJ$)

During the free expansion of an ideal gas,its.......