1

### NEET 2016 Phase 2

For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by
A
$\Delta S = nR\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)$
B
$\Delta S = nR\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)$
C
$\Delta S = nRT\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)$
D
$\Delta S = RT\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)$

## Explanation

For an ideal gas undergoing reversible expansion, when temperature changes from Ti to Tf and pressure changes from Pi to Pf.

$\Delta$S = nCp ln ${{{T_f}} \over {{T_i}}}$ + nR ln ${{{P_i}} \over {{P_f}}}$

For an isothermal process, Ti = Tf so, ln 1 = 0

$\therefore$ $\Delta$S = nR ln ${{{P_i}} \over {{P_f}}}$
2

### AIPMT 2015

The heat of combination of carbon to CO2 is $-$393.5 kJ/mol. The heat released upon formation of 35.2 g of CO2 from carbon and oxygen gas is
A
+ 315 kJ
B
$-$630 kJ
C
$-$ 3.15 kJ
D
$-$ 315 kJ

## Explanation

Given
${C_{(s)}} + {O_{2(g)}} \to C{O_{(g)}},\Delta H = - 393.5\,kJ/mol$

Then amount of heat released on formation of 44 g CO2 = 393.5 kJ

$\therefore$ Amount of heat released on formation of

35.2 g CO2 = ${{393.5} \over {44}} \times 35.2 = 314.8 \approx 315\,kJ$
3

### AIPMT 2014

For the reaction, ${X_2}{O_{4\left( l \right)}}\,\, \to \,\,2X{O_{2(g)}}$
$\Delta$U = 2.1 kcal, $\Delta$S = 20 cal K$-$1 at 300 K
A
2.7 kcal
B
$-$ 2.7 kcal
C
9.3 kcal
D
$-$ 9.3 kcal

## Explanation

$\Delta H = \Delta U + \Delta {n_g}RT$

Given $\Delta U = 2.1\,kcal,\,\Delta {n_g} = 2$

$R = 2 \times {10^{ - 3}}kcal,T = 300K$

$\therefore$ $\Delta H = 2.1 + 2 \times 2 \times {10^{ - 3}} \times 300 = 3.3\,kcal$

Again, $\Delta G = \Delta H + T\Delta S$

Given $\Delta$S = 20 $\times$ 10-3 kcal K-1

On putting the values of $\Delta$H and $\Delta$S in the equation, we get

$\Delta G = 3.3 - 300 \times 20 \times {10^{ - 3}}$

$\Rightarrow 3.3 - 6 \times {10^3} \times {10^{ - 3}} = - 2.7\,kcal$
4

### AIPMT 2014

Which of the following statements is correct for the spontaneous adsorption of a gas?
A
$\Delta$S is negative and, therefore $\Delta$H should be highly positive.
B
$\Delta$S is negative and therefore, $\Delta$H should be highly negative.
C
$\Delta$S is positive and therefore, $\Delta$H should be negative .
D
$\Delta$S is positive and therefore, $\Delta$H should also be highly positive.

## Explanation

Using Gibb's-Helmholtz equation,

$\Delta G = \Delta H - T\Delta S$

During adsorption of a gas, entropy decreases i.e, $\Delta$S < 0

For spontaneous adsorption, $\Delta$G should be negative, which is possible when $\Delta$H is highly negative.