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1

AIPMT 2003

MCQ (Single Correct Answer)
The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm$$-$$3, respectively. If the standard free energy difference $$\left( {\Delta {G^o}} \right)$$ is equal to 1895 J mol$$-$$1, the pressure at which graphite will be transformed into diamond at 298 K is
A
11.14 $$ \times $$ 108 Pa
B
11.14 $$ \times $$ 107 Pa
C
11.14 $$ \times $$ 106 Pa
D
11.14 $$ \times $$ 105 Pa

Explanation

C(graphite) $$ \to $$ C(diamond)

Volume of graphite = $${{12} \over {2.25}} = 5.33$$ cm3 mol–1

Volume of diamond = $${{12} \over {3.31}} = 3.63$$ cm3 mol–1

$$\Delta $$V = Vgraphite – Vdiamond

= 1.70 cm3 mol–1

= 1.70 × 10–3 L mol–1

So, $$\Delta $$Go = P$$\Delta $$V

$$ \Rightarrow $$ 1895 J mol–1 = P(1.70 × 10–3 L mol–1)

$$ \Rightarrow $$ 1114.70 × 103 J/L = P

$$ \Rightarrow $$ $${{1114.7 \times {{10}^3}} \over {101.33}}$$ = P

$$ \Rightarrow $$ P = 11000.69 atm × 1.013 × 105 Pa

= 11143.69 × 105 Pa

= 11.14 × 108 Pa
2

AIPMT 2003

MCQ (Single Correct Answer)
For the reaction,
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
at constant temperature, $$\Delta $$H $$-$$ $$\Delta $$E is
A
+ RT
B
$$-$$3RT
C
+ 3RT
D
$$-$$ RT

Explanation

$$\Delta $$H = $$\Delta $$E + $$\Delta $$ng RT

$$\Delta $$ng = 3 – (1 + 5) = – 3

$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$E + (– 3RT)

$$ \Rightarrow $$ $$\Delta $$H – $$\Delta $$E = – 3RT
3

AIPMT 2002

MCQ (Single Correct Answer)
2 mole of ideal gas at 27oC temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)
A
92.1
B
0
C
4
D
9.2

Explanation

For isothermal reversible expansion

w = q = nRT $$ \times $$ 2.303 $$\log {{{v_2}} \over {{v_1}}}$$

= 2RT $$ \times $$ 2.303 $$\log {{{20}} \over {{2}}}$$

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, $$\Delta $$S = $${q \over T}$$ = $${{9.2T} \over T}$$ = 9.2 cal/mol K
4

AIPMT 2002

MCQ (Single Correct Answer)
Heat of combustion $$\Delta $$Ho for C(s), H2(g) and CH4(g) are $$-$$ 94, $$-$$ 68 and $$-$$213 kcal/mol, then $$\Delta $$Ho for C(s) + 2H2(g) $$ \to $$ CH4(g) is
A
$$-$$17 kcal
B
$$-$$111 kcal
C
$$-$$170 kcal
D
$$-$$85 kcal

Explanation

C(s) + 2H2(g) $$ \to $$ CH4(g), $$\Delta $$Ho = ?

C(s) + O2(g) → CO2(g), $$\Delta $$H = – 94 kcal/mol ....(1)

H2(g) + $${1 \over 2}$$O2(g) $$ \to $$ H2O(g), $$\Delta $$H = – 68 kcal/mol ....(2)

CH4(g) + 2O2(g) $$ \to $$ CO2(g) + H2O(l), $$\Delta $$H = – 213 kcal/mol ....(3)

Performing (1) + 2 × (2) – (3)

$$\Delta $$Ho = [– 94 – 2(68)] – (–213) kcal/mol

= [– 94 – 136] + 213 k cal/mol

= – 230 + 213 = – 17 kcal/mol

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