The correct statement among the following is:

1.	The presence of reacting species in a covered beaker is an example of an open system.
2.	There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
3.	The presence of reactants in a closed vessel made up of copper is an example of a closed system.
4.	The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

If the volume of a gas is reduced to half from its original volume, then the specific heat will:

During the complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for this change is:

${∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$ of formation of ${\mathrm{CH}}_4\left(\mathrm{g}\right)$ at certain temperature is $-393$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$. The value of ${∆}_{\mathrm{f}}{\mathrm{H}}^{⊝}$ is-

1. Zero

2. $<{∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

3. $>{∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

4. Equal to ${∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

In an adiabatic process, no transfer of heat takes place between the system and its surroundings. The correct option for free expansion of an ideal gas under adiabatic condition from the following is:

1. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}\ne 0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

2. $\mathrm{q}\ne 0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

3. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

4. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}\ne 0$

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:

On the basis of thermochemical equations (i), (ii), and (iii), find out which of the algebraic relationships given in options (a) to (d) is correct:

1. z=x+y

2. x=y-z

3. x=y+z

4. y=2z-x

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-