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.