1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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.

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