$A$ solution of $500 \ mL$ of $2 \ M \ KOH$ is added to $500 \ mL$ of $2 \ M \ HCl$ and the mixture is well shaken. The rise in temperature $T_1$ is noted. The experiment is then repeated using $250 \ mL$ of each solution and the rise in temperature $T_2$ is again noted. Assume all heat is absorbed by the solution.

$A$ calorimeter contains $0.2 \ kg$ of water at $30^{\circ} C$. $0.1 \ kg$ of water at $60^{\circ} C$ is added to it,the mixture is well stirred and the resulting temperature is found to be $35^{\circ} C$. The water equivalent of the calorimeter is $:-$ (in $J / K$)

$9.0 \ g$ of $H_2O$ is vaporised at $100 \ ^oC$ and $1 \ atm$ pressure. If the latent heat of vaporisation of water is $x \ J/g$,then $\Delta S$ is given by

The value of $\Delta H_{transition}$ for $C(\text{graphite}) \rightarrow C(\text{diamond})$ is $1.9 \ kJ/mol$ at $25^{\circ}C$. The entropy of graphite is higher than the entropy of diamond. This implies that which of the following is incorrect $:-$

An ideal gas in a thermally insulated vessel at internal pressure $= P_1$,volume $= V_1$ and absolute temperature $= T_1$ expands irreversibly against zero external pressure,as shown in the diagram. The final internal pressure,volume and absolute temperature of the gas are $P_2, V_2$ and $T_2$,respectively. For this expansion,
$(A) \ q = 0$
$(B) \ T_2 = T_1$
$(C) \ P_2 V_2 = P_1 V_1$
$(D) \ P_2 V_2^\gamma = P_1 V_1^\gamma$

$1 \ g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 \ K$ and $1 \ atm$ atmospheric pressure according to the equation:
$C \ (graphite) + O_{2(g)} \rightarrow CO_{2(g)}$
During the reaction,the temperature rises from $298 \ K$ to $299 \ K$. If the heat capacity of the bomb calorimeter is $20.7 \ kJ \ K^{-1}$,what is the enthalpy change for the above reaction at $298 \ K$ and $1 \ atm$?