Embibe Experts Solutions for Chapter: Thermodynamics, Exercise 3: Exercise-3

An ideal gas in a thermally insulated vessel at internal pressure $={\mathrm{P}}_1,$ volume $={\mathrm{V}}_1$ and absolute temperature $={\mathrm{T}}_1$ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are ${\mathrm{P}}_2, {\mathrm{V}}_2$ and ${\mathrm{T}}_2$, respectively. For this expansion

A piston filled with $0.04$ mol of an ideal gas expands reversibly from $50.0 \mathrm{mL}$ to $375 \mathrm{mL}$ at a constant temperature of $37.0°\mathrm{C}.$ As it does so, it absorbs $208 \mathrm{J}$ of heat. The values of $\mathrm{q}$ and $\mathrm{w}$ for the process will be:

$(\mathrm{R}=8.314 \mathrm{J} {\mathrm{mol}}^{-1} \mathrm{K})$ $(\ln 7.5=2.01)$

$(\mathrm{R}=8.314 \mathrm{J} {\mathrm{mol}}^{-1} \mathrm{K}) (\ln 7.5=2.01)$

$△\mathrm{U}$ equal to:

For an ideal gas, consider only $\mathrm{P}-\mathrm{V}$ work in going from an initial state $\mathrm{X}$ to the final state $\mathrm{Z}.$ The final state $\mathrm{Z}$ can be reached by either of the two paths shown in the figure. Which of the following choice$\left(\mathrm{S}\right)$ is (are) correct? [take $\mathrm{\Delta S}$ as change in entropy and w as work done].

Identify the CORRECT statement regarding a spontaneous process.

In conversion of lime-stone to lime, ${\mathrm{CaCO}}_{3\left(\mathrm{s}\right)}⟶{\mathrm{CaO}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$ the values $△\mathrm{H}°$ and $△\mathrm{S}°$ are $179.1 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $160.2 \mathrm{J} {\mathrm{K}}^{-1}-{\mathrm{mol}}^{-1}$ respectively at $298 \mathrm{K}$ and $1 \mathrm{bar}$. Assuming that $△\mathrm{H}°$ and $△\mathrm{S}°$ do not change with temperature, temperature above which conversion of lime stone to lime will be spontaneous is :

The heats of combustion of carbon and carbon monoxide are $-393.5$ and $-283.5 \mathrm{kJ} {\mathrm{mol}}^{-1},$ respectively. The heat of formation (in $\mathrm{kJ}$) of carbon monoxide per mole is :

The value of ${\log }_{10}\mathrm{K}$ for a reaction $\mathrm{A}\rightleftharpoons \mathrm{B}$ is

(Given ${\mathrm{\Delta }}_{\mathrm{r}}{{\mathrm{H}}^{\mathrm{o}}}_{298\mathrm{K}}=-54.07 {\mathrm{kJmol}}^{-1}, {\mathrm{\Delta }}_{\mathrm{r}}{{\mathrm{S}}^{\mathrm{o}}}_{298\mathrm{K}}=10 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ and $\mathrm{R}=8.314 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}; 2.303\times 8.314\times 298=5705$)