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1

### AIPMT 2012 Prelims

Standard enthalpy of vaporisation $$\Delta$$vapHo for water at 100oC is 40.66 kJ mol$$-$$1. The internal energy of vaporisation of water at 100oC (in kJ mol$$-$$1) is
A
+37.56
B
$$-$$43.76
C
+ 43.76
D
+ 40.66

## Explanation

H2O(l) $$\buildrel {100^\circ \,C} \over \longrightarrow$$ H2O(g)

∆Ho = 40.66kJ mol–1

∆Ho = ∆uo + $$\Delta$$ng RT

$$\Delta$$ng = 1, R = 8.314 × 10–3 kJ mol–1 k–1

T = 100 + 273 = 373 K

$$\Rightarrow$$ 40.66 = ∆uo + (1) (8.314 × 10–3) × 373

$$\Rightarrow$$ ∆uo = 37.56 kJ mol–1
2

### AIPMT 2012 Prelims

In which of the following reactions, standard reaction entropy change ($$\Delta$$So) is positive and standard Gibb's energy change ($$\Delta$$Go) decreases sharply with increasing temperature ?
A
C(graphite) + $${1 \over 2}$$O2(g) $$\to$$ CO(g)
B
CO(g) + $${1 \over 2}$$O2(g) $$\to$$ CO2(g)
C
Mg(s) + $${1 \over 2}$$O2(g) $$\to$$ MgO(g)
D
$${1 \over 2}$$C(graphite) + $${1 \over 2}$$O2(g) $$\to$$ $${1 \over 2}$$CO2(g)

## Explanation

$$\Delta$$So positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the $$\Delta$$So is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O2 and Carbon (C) in solid state. Thus, reactions have less randomness than products. Also, the reaction is a combustion reaction which has a negative value of $$\Delta$$H. According to Gibbs energy change ($$\Delta$$So).

$$\Delta$$Go = $$\Delta$$Ho - T$$\Delta$$So

$$\Rightarrow$$ $$\Delta$$Go = -ve - T(+ve)

Thus, on increasing temperature value of $$\Delta$$Go decreases sharply.
3

### AIPMT 2011 Mains

Consider the following processes :

H (kJ/mol)
1/2A $$\to$$ B +150
3B  $$\to$$ 2C + D -125
E + A  $$\to$$ 2D +350

For B + D $$\to$$ E + 2C, $$\Delta$$H will be
A
525 kJ/mol
B
$$-$$175 kJ/mol
C
$$-$$ 325 kJ/mol
D
325 kJ/mol

## Explanation

H (kJ/mol)
1/2A $$\to$$ B +150
3B  $$\to$$ 2C + D -125
E + A  $$\to$$ 2D +350
___________________________________
B + D $$\to$$ E + 2C;

$$\Delta$$H = (300 - 125 - 350) = - 175 kJ/mol
4

### AIPMT 2011 Prelims

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?
A
q = 0, $$\Delta$$T $$\ne$$ 0, w = 0
B
q $$\ne$$ 0,   $$\Delta$$T = 0,   w = 0
C
q = 0,   $$\Delta$$T = 0,   w = 0
D
q = 0,  $$\Delta$$T < 0, w $$\ne$$ 0

## Explanation

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $$\Delta$$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.

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