$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

State $1 \longleftarrow$ State $2 \longleftarrow$ State $3$
$\left(\begin{array}{c} T=300 \ K \\ P=15 \ bar \\ 1 \ mole \end{array}\right) \left(\begin{array}{c} T=300 \ K \\ P=10 \ bar \\ 1 \ mole \end{array}\right) \left(\begin{array}{c} T=300 \ K \\ P=5 \ bar \\ 1 \ mole \end{array}\right)$
The above shows a cyclic process. Calculate the total work done during one complete cycle. (Assume a single step to reach the next state).

Arrange the following isothermal processes in order of the magnitude of the work $(w)$ involved between states $1$ and $2$.
$A$. Expansion in single stage $(w_A)$
$B$. Expansion in multi stages $(w_B)$
$C$. Compression in single stage $(w_C)$
$D$. Compression in multi stages $(w_D)$

Assertion : $\Delta H$ and $\Delta E$ are almost same for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$.
Reason : All reactants and products are gases.

The heat of reaction for $C_2H_2 + H_2 \rightarrow C_2H_4$ is given by the following data:
$(i) \Delta H_f^o \text{ of } H_2O_{(\ell)} = -68.3 \ K \ cal \ mol^{-1}$
$(ii) \Delta H_{comb}^o \text{ of } C_2H_2 = -337.2 \ K \ cal \ mol^{-1}$
$(iii) \Delta H_{comb}^o \text{ of } C_2H_4 = -363.7 \ K \ cal \ mol^{-1}$

Molar enthalpy change for vapourisation of $1.0 \ mol$ of water at $1.0 \ bar$ and $100 ^{\circ} C$ is $41.0 \ kJ \ mol^{-1}$. If water vapour is assumed to be an ideal gas,the internal energy change for $1.0 \ g$ of water in $kJ$ is