One mole of an ideal monoatomic gas at temperature $T$ expands adiabatically in an insulated flask from $1 \ L$ to $2 \ L$ against an external pressure of $1 \ atm$. What is the final temperature?

Enthalpy of fusion and enthalpy of vaporization for water respectively are $6.01 \ kJ \ mol^{-1}$ and $45.07 \ kJ \ mol^{-1}$ at $0^{\circ}C$. What is enthalpy of sublimation at $0^{\circ}C$?

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.
Reason : For all isothermal processes,$\Delta E = 0$.

One mole of an ideal gas at $900 \ K$ undergoes two reversible processes,$I$ followed by $II$,as shown in the graph. If the work done by the gas in the two processes is the same,the value of $\ln \frac{V_3}{V_2}$ is. . . . . . . . ($U$: internal energy,$S$: entropy,$p$: pressure,$V$: volume,$R$: gas constant). (Given: molar heat capacity at constant volume,$C_{V, m}$ of the gas is $\frac{5}{2} R$)

The $\Delta H$ and $\Delta S$ for a reaction at $1 \ \text{atm}$ pressure are $+30.558 \ \text{kJ}$ and $0.066 \ \text{kJ K}^{-1}$ respectively. The temperature at which the free energy change will be zero and below this temperature the nature of the reaction would be:

Considering the reaction $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} + 393.5 \ kJ$,the signs of $\Delta H, \Delta S$,and $\Delta G$ respectively are: