At the temperature $T$ $(K)$ for the reaction: $X_2O_{4(l)} \rightarrow 2XO_{2(g)}$,given $\Delta U = x \ kJ \ mol^{-1}$ and $\Delta S = y \ J \ K^{-1} \ mol^{-1}$. The Gibbs energy change for the reaction is: (Assume $X_2O_4$ and $XO_2$ are ideal gases)

One mole of an ideal monoatomic gas undergoes two reversible processes ($A \rightarrow B$ and $B \rightarrow C$) as shown in the given figure:
$A \rightarrow B$ is an adiabatic process. If the total heat absorbed in the entire process ($A \rightarrow B$ and $B \rightarrow C$) is $R T_2 \ln 10$,the value of $2 \log V_3$ is . . . . . [Use,molar heat capacity of the gas at constant pressure,$C_{p, m} = \frac{5}{2} R$ ]

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,$\Delta E = 0$.

$20.0 \ dm^{3}$ of an ideal gas '$X$' at $600 \ K$ and $0.5 \ MPa$ undergoes isothermal reversible expansion until the pressure of the gas is $0.2 \ MPa$. Which of the following options is correct? (Given: $\log\,2=0.3010$ and $\log\,5=0.6989$)

Which of the following is/are true about the reversible isothermal expansion of an ideal gas?
$(a) \ \Delta U = 0$
$(b) \ q = 0$
$(c) \ \Delta T = 0$
$(d) \ q = 2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$

The reaction $NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + CO_{2(g)} + H_2O_{(l)}$ is carried out in a bomb calorimeter. The heat evolved is $743 \ kJ \ mol^{-1}$. Calculate the value of $\Delta H$ at $300 \ K$ in $kJ \ mol^{-1}$.