The ratio of ${\mathrm{C}}_{\mathrm{p}}$ value of a triatomic gas to the ${\mathrm{C}}_{\mathrm{v}}$ value of a monatomic gas is

One mole of an ideal gas is taken from $\mathrm{a}\to \mathrm{b}$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ${\mathrm{W}}_{\mathrm{s}}$ and that along the dotted line path is ${\mathrm{W}}_{\mathrm{d}}$, then the integer closest to the ratio $\frac{{\mathrm{W}}_{\mathrm{d}}}{{\mathrm{W}}_{\mathrm{s}}}$ is

The surface of copper gets tarnished by the formation of copper oxide. ${\mathrm{N}}_2$ gas was passed to prevent the oxide formation during heating of copper at $1250 \mathrm{K}$. However, the${\mathrm{N}}_2$ gas contains $1 \mathrm{mole} \%$ of water vapour as impurity. The water vapour oxidise copper as per the reaction given below :

$2 \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Cu}}_2\mathrm{O}+{\mathrm{H}}_2\left(\mathrm{g}\right)$

${\mathrm{P}}_{{\mathrm{H}}_2}$ is the minimum partial pressure of ${\mathrm{H}}_2$ (in bar) needed to prevent the oxidation at $1250 \mathrm{K}$. The value of

$\ln \left({\mathrm{P}}_{{\mathrm{H}}_2}\right)$ is...........

(Given : total pressure= $1 \mathrm{bar}$, $\mathrm{R}$ (universal gas constant) = $8 {\mathrm{JK}}^{-1},{\mathrm{mol}}^{-1},$ $\ln \left(10\right)=2.3$, $\mathrm{Cu} \left(\mathrm{s}\right)$ and ${\mathrm{Cu}}_2\mathrm{O} \left(\mathrm{s}\right)$are mutually immiscible

At $1250 \mathrm{K}$:  $2 \mathrm{Cu} \left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2 \left(\mathrm{g}\right)\to {\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{s}\right), ∆\mathrm{G}°=-78,000 \mathrm{J} {\mathrm{mol}}^{-1}$

${\mathrm{H}}_2 \left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2 \left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right),∆\mathrm{G}°=-1,78000 \mathrm{J} {\mathrm{mol}}^{-1}$

$\mathrm{G}$ is Gibbs energy)

$500 \mathrm{ml} \mathrm{of} 0.1 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$was added into $1\mathrm{L} \mathrm{of} 0.1\mathrm{M} \mathrm{NaOH}$ solution. The heat evolved was $\mathrm{x}$ calories. If further $500 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4$is added into the solution, now heat evolved will be

Calculate the enthalpy of formation of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ from the following data :-

${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}+12{\mathrm{O}}_2\to 12{\mathrm{CO}}_2+11{\mathrm{H}}_2\mathrm{O}$, $\mathrm{\Delta H}=–5200.7 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

$\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2, \mathrm{\Delta H}=–394.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}, \mathrm{\Delta H}=–285.8 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

Calculate the enthalpy of reaction for $\mathrm{CO}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)$.

Given,

$\mathrm{C}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right), ∆\mathrm{H}=-393.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $\mathrm{C}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{CO}\left(\mathrm{g}\right), ∆\mathrm{H}=-110.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$.