At $300 \ K$, $3.0 \ \text{moles}$ of an ideal gas at $3.0 \ \text{atm}$ pressure is compressed isothermally to one half of its volume by an external pressure of $6.0 \ \text{atm}$. The work done (in $kJ$) is. Given, $R=0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$ $(1 \ \text{L atm} = 101.3 \ \text{J})$.

The volume of a gas decreases from $500 \ cc$ to $300 \ cc$ when a sample of gas is compressed by an average pressure of $0.6 \ atm$. During this process $10 \ J$ of heat is liberated. The change in internal energy is......$J$

For the reactions,$C + O_2 \to CO_2; \Delta H = -393 \ J$
$2Zn + O_2 \to 2ZnO; \Delta H = -412 \ J$

For the reaction: $A_{(g)} + B_{(s)} \rightleftharpoons 2C_{(g)} + D_{(g)}$,given $\Delta U = 5.0 \ kcal$ and $\Delta S = 50 \ cal \ K^{-1}$ at $400 \ K$. Calculate $\Delta G$ for the reaction. (in $kcal$)

In a constant volume calorimeter,$3.5 \ g$ of a gas with molecular weight $28$ was burnt in excess oxygen at $298.0 \ K$. The temperature of the calorimeter was found to increase from $298.0 \ K$ to $298.45 \ K$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 \ kJ \ K^{-1}$,the numerical value for the enthalpy of combustion of the gas in $kJ \ mol^{-1}$ is

The enthalpy change for the reaction of $50.00 \ mL$ of ethylene with $50.00 \ mL$ of $H_2$ at $1.5 \ atm$ pressure is $\Delta H = -0.31 \ kJ$. The value of $\Delta E$ will be (in $kJ$)