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

An ideal gas expanded irreversibly against $10 \ bar$ pressure from $20 \ L$ to $30 \ L$. Calculate $Q$ if the process is isoenthalpic. $(1 \ L \ bar = 100 \ J)$

Consider the following statements:
Statement-$I$: Both internal energy $(U)$ and work $(w)$ are state functions.
Statement-$II$: During the free expansion of an ideal gas into vacuum,the work done is zero.
The correct answer is:

Standard entropies of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightleftharpoons XY_3, \Delta H = -30 \ kJ$,to be at equilibrium,the temperature should be ............. $K$.

Consider a reaction $2H_2O_{(l)} \to 2H_{2(g)} + O_{2(g)}$. Calculate the work done at $25\, ^oC$ for the decomposition of $36\, mL$ of water. (in $, KCal$)

The molar heat capacity $(C_p)$ of $CD_2O$ is $10 \, cal \, K^{-1} \, mol^{-1}$ at $1000 \, K$. The change in entropy associated with cooling of $32 \, g$ of $CD_2O$ vapour from $1000 \, K$ to $100 \, K$ at constant pressure will be.....$cal \, deg^{-1}$ ($D = $ deuterium,atomic mass $= 2 \, u$)