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1

### 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

### 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

### 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

### 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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