The pressure-volume work for an ideal gas can be calculated by using the expression $\mathrm{W}={\int }_{{\mathrm{V}}_{\mathrm{i}}}^{{\mathrm{V}}_{\mathrm{f}}}{\mathrm{p}}_{\mathrm{ex}}d\mathrm{V}$.

The work can also be calculated from the pV-plot by using the area under the curve within the specified limits.
An ideal gas is compressed (a) reversibly or (b) irreversibly from volume ${\mathrm{V}}_{\mathrm{i}}$ to ${\mathrm{V}}_{\mathrm{f}}$.
The correct option is:

1. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)=\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{irreversible}\right)$

2. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)<\mathrm{W}\hspace{0.17em}\left(\mathrm{irreversible}\right)$

3. $\mathrm{W}<mspace\; width="1em"></mspace>(\mathrm{reversible})>\mathrm{W}<mspace\; width="1em"></mspace>(\mathrm{irreversible})$

4. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)=\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{irreversible}\right)+{\mathrm{p}}_{\mathrm{ex}}.∆\mathrm{V}$

The entropy change can be calculated by using the expression $∆\mathrm{S}=\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$. When water freezes in a glass beaker, the correct statement among the following is:

Based on the reactions, ascertain the valid algebraic relationship:
$\left(i\right)$ $C\left(g\right)$ $+4H$ $\left(g\right)$ $\to C{H}_4\left(g\right);$ $$
$∆H_r$ $=x$ $kJ$ $mo{l}^{-1}$
$\left(ii\right)$ $C\left(graphite\right)$ $+$ $2H_2$ $\left(g\right)$ $\to$ $$ $C{H}_4;$ $$
$∆H_r$ $$ $=$ $y$ $kJ$ $mo{l}^{-1}$

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound is-

Consider the following statement:

"A thermodynamic quantity is a state function"

The correct option is-

1. Used to determine heat changes

2. Whose value is independent of the path

3. Used to determine pressure-volume work

4. Whose value depends on temperature only.

What is a necessary condition for an adiabatic process to occur?

1. ∆T = 0

2. ∆P = 0

3. q = 0

4. w = 0

The enthalpy of formation of all elements in their standard state is-

$∆\mathrm{U}°$ $$for combustion of methane is –x kJ mol^–1.

The value of $∆\mathrm{H}°$ for the same reaction would be:

$1.$ $=$ $∆\mathrm{U}°$
$2.$ $>$ $∆\mathrm{U}°$
$3.$ $<$ $∆\mathrm{U}°$
$4.$ $=0$

701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy for the process is:

The reaction of cyanamide, ${\mathrm{NH}}_2\mathrm{CN}$ $\left(\mathrm{s}\right)$ with dioxygen, was carried out in a bomb calorimeter, and ∆U was found to be $-742.7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$ at 298 K.
\(\small{\mathrm{NH}_2 \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{g})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})}\)

${}_{}$The enthalpy change for the reaction at 298 K would be:

$1.$ $-741.3$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $+$ $753.9$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$3.$ $+$ $772.$ $7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-845.$ $1$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$