1

### NEET 2017

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy $\Delta$U of the gas in joules will be
A
$-$ 500 J
B
$-$ 505 J
C
+ 505 J
D
1136.25 J

## Explanation

w = - Pext$\Delta$V = -2.5(4.50 - 2.50)

$\Rightarrow$ - 5 L atm = - 5 $\times$ 1.01.325 J = - 506.625 J

$\Delta$U = q + w

As, the container is insulted, thus q = 0

Hence, $\Delta$U = w = -506.625 J
2

### NEET 2017

For a given reaction, $\Delta$H = 35.5 kJ mol$-$1 and $\Delta$S = 83.6 J K$-$1 mol$-$1. The reaction is spontaneous at (Assume that $\Delta$H and $\Delta$S do not vary with temperature.)
A
T > 425 K
B
all temperatures
C
T > 298 K
D
T < 425 K

## Explanation

For a spontaneous reaction,

$\Delta$G < 0 i.e. $\Delta$H - T$\Delta$S < 0

$T > {{\Delta H} \over {\Delta S}}$

$T > \left( {{{35.5 \times 1000} \over {83.6}} = 424.6 \approx 425K} \right)$

$\therefore T > 425\,K$
3

### 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 ${{{P_i}} \over {{P_f}}}$
4
MCQ (More than One Correct Answer)

### NEET 2016 Phase 1

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
A
$\Delta$H < 0 and $\Delta$S > 0
B
$\Delta$H < 0 and $\Delta$S < 0
C
$\Delta$H < 0 and $\Delta$S = 0
D
$\Delta$H > 0 and $\Delta$S < 0

## Explanation

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

If $\Delta$H < 0 and $\Delta$S > 0
$\Delta G = (-ve) - T (+ve)$

then at all temperatures, $\Delta G$ = -ve, spontaneous reaction.

If $\Delta$H < 0 and $\Delta$S = 0
$\Delta G = (-ve) - T (0)$

then at all temperatures, $\Delta G$ = -ve at all temperatures.