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

For the reaction,$2Cl_{(g)} \to Cl_{2(g)}$,the signs of $\Delta H$ and $\Delta S$ respectively,are:

How much heat in $KJ$ is released during the combustion of $112 \ L$ of water gas at $S.T.P.$? (Water gas is an equimolar mixture of $H_2$ and $CO$)
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}; \Delta H = -241.8 \ KJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -283 \ KJ$

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

In an adiabatic process,the work involved during expansion or compression of an ideal gas is given by

$28.0 \, L$ of $CO_2$ is produced on complete combustion of $16.8 \, L$ gaseous mixture of ethene $(C_2H_4)$ and methane $(CH_4)$ at $25^{\circ}C$ and $1 \, atm$. Heat evolved during the combustion process is $......... \, kJ$.
Given :
$\Delta H_C(CH_4) = -900 \, kJ \, mol^{-1}$
$\Delta H_C(C_2H_4) = -1400 \, kJ \, mol^{-1}$