1
MCQ (Single Correct Answer)

AIPMT 2002

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
2
MCQ (Single Correct Answer)

AIPMT 2002

Which reaction is not feasible?
A
2KI + Br2 $$ \to $$ 2KBr + I2
B
2KBr + I2 $$ \to $$ 2KI + Br2
C
2KBr + Cl2 $$ \to $$ 2KCl + Br2
D
2H2O + 2F2 $$ \to $$ 4HF + O2

Explanation

2KBr + I2 $$ \to $$ 2KI + Br2

reaction is not possible because Br ion is not oxidised in Br2 with I2 due to higher electrode (oxidation) potential of I2 than bromine.
3
MCQ (Single Correct Answer)

AIPMT 2002

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
4
MCQ (Single Correct Answer)

AIPMT 2002

In a closed insulated container a liquid is stirred with a paddle to increase the temperature which of the following is true ?
A
$$\Delta $$E = W $$ \ne $$ 0, q = 0
B
$$\Delta $$E = w = q $$ \ne $$ 0
C
$$\Delta $$E = 0, W = q $$ \ne $$ 0
D
W = 0, $$\Delta $$E = q $$ \ne $$ 0.

Explanation

When in a closed insulated container, a liquid is stirred, it increases the temperature of liquid thus it started behaving as adiabatic process resulting no change in heat and only temperature increases. Therefore, q = 0 for this process.

Also, $$\Delta $$E = q + W

$$\Delta $$E = W (As, q = 0)

As, there is rise in temperature thus, there must be change in internal energy, i.e., $$\Delta $$E = W ≠ 0.

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