Consider the $P-V$ (pressure-volume) diagram given below,where an ideal gas is reversibly converted from state $X$ to state $Y$. Among the following,which is the correct $T-S$ (temperature-entropy) diagram corresponding to this process?

The molar enthalpy change for $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $373 \ K$ and $1 \ atm$ is $41 \ kJ \ mol^{-1}$. Assuming ideal behavior,the internal energy change for vaporization of $1 \ mol$ of water at $373 \ K$ and $1 \ atm$ in $kJ \ mol^{-1}$ is:

Given
$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)};$
$\Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)};$
$\Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)};$
$\Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
Based on the above thermochemical equations,the value of $\Delta_rH^o$ at $298 \, K$ for the reaction
$C_{(graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)}$ will be ........... $kJ \, mol^{-1}$.

$1 \, \text{mole}$ of $H_2SO_4$ is mixed with $2 \, \text{moles}$ of $NaOH$. The heat evolved will be

The standard entropies of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $X_2 + 3Y_2 \rightleftharpoons 2XY_3; \Delta H = -60 \ kJ$ to be at equilibrium,the temperature should be.....$K$

If $\Delta G$ and $\Delta S$ for the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$ at $2000 \ K$ are $-40 \ kJ \ mol^{-1}$ and $0.22 \ kJ \ K^{-1} \ mol^{-1}$ respectively,the change in internal energy for the same reaction approximately (in $kJ \ mol^{-1}$) is