If a refrigerator's door is opened in a closed room,then we get:

Total enthalpy change for freezing of $1 \ mol$ of water at $10^{\circ} C$ to ice at $-10^{\circ} C$ is $..........$ (Given : $\Delta_{fus} H = x \ kJ / mol$,$C_{p}[H_2O_{(l)}] = y \ J \ mol^{-1} \ K^{-1}$,$C_{p}[H_2O_{(s)}] = z \ J \ mol^{-1} \ K^{-1}$)

For the reaction,$2 CO + O_2 \longrightarrow 2 CO_2$; $\Delta H = -560 \ kJ$. Two moles of $CO$ and one mole of $O_2$ are taken in a container of volume $1 \ L$. They completely form two moles of $CO_2$. The gases deviate appreciably from ideal behavior. If the pressure in the vessel changes from $70 \ atm$ to $40 \ atm$,find the magnitude (absolute value) of $\Delta U$ at $500 \ K$. $(1 \ L \ atm = 0.1 \ kJ)$

Identify the incorrect statements from the following.
$I$. For adiabatic process,$\Delta U = w_{ad}$
$II$. Enthalpy is an intensive property
$III$. For the process,$H_2O_{(l)} \rightarrow H_2O_{(s)}$,the entropy increases

$A$ sample of argon at $1 \text{ atm}$ pressure and $300 \text{ K}$ expands reversibly and adiabatically from $1.25 \text{ dm}^3$ to $2.5 \text{ dm}^3$. Calculate the approximate enthalpy change (in $\text{J}$).
$(I)$ $C_V$ for argon is $12.48 \text{ J K}^{-1} \text{ mol}^{-1}$
$(II)$ Assume argon to be an ideal gas
$(III)$ $\Delta T = 111.5 \text{ K}$ (temperature decrease)

$CsOH + HCl \to CsCl + H_2O$,$\Delta H = -13.4 \ K \ cal/mol$
$CsOH + HA \to CsA + H_2O$,$\Delta H = -10.4 \ K \ cal/mol$
Then calculate $\Delta H$ of ionisation of $HA$ in $K \ cal/mol$.