The temperature of $1 \ mol$ of an ideal gas is increased by $2 \ ^oC$ at constant pressure. The work done is:

Two liters of $N_2$ gas at $0 \, ^\circ C$ and $5 \, atm$ pressure undergoes isothermal expansion against a constant external pressure of $1 \, atm$ until the pressure of the gas becomes $1 \, atm$. If the gas is ideal,what is the work done in the expansion in $J$?

The difference between the reaction enthalpy change $(\Delta _r H)$ and reaction internal energy change $(\Delta _r U)$ for the reaction $2C_6H_{6(l)} + 15O_{2(g)} \longrightarrow 12CO_{2(g)} + 6H_2O_{(l)}$ at $300 \ K$ is $....$ $J \ mol^{-1}$ $(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

The energy of combustion per mole of $H_2$,$LPG$,and octane follows the order:

Match the following processes with their corresponding entropy changes:

Process	Entropy Change
$(a)$ Liquid to vapor conversion	$(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature	$(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas	$(3)$ $\Delta S = (-)$

Enthalpy of formation of $CO_{2(g)}$,$H_2O_{(l)}$ and $C_6H_{12}O_{6(s)}$ are $-393$,$-286$ and $-1170 \ kJ \ mol^{-1}$ respectively. The quantity of heat liberated when $18 \ g$ of $C_6H_{12}O_{6(s)}$ is burnt completely in oxygen is (in $kJ$)