1
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.
2
MCQ (Single Correct Answer)

### AIPMT 2002

Unit of entropy is
A
J K$-$1 mol$-$1
B
J mol$-$1
C
J$-$1 K$-$1 mol$-$1
D
J K mol$-$1

## Explanation

Entropy is the change in heat per degree. Thus, its unit is as follows.

Entropy = ${{Jmo{l^{ - 1}}} \over K}$ = J mol-1 K-1
3
MCQ (Single Correct Answer)

### AIPMT 2001

Enthalpy of CH4 + ${1 \over 2}$ O2 $\to$ CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x and y respectively. Then which relation is correct?
A
x > y
B
x < y
C
x = y
D
x3 y

## Explanation

CH4 + ${1 \over 2}$ O2 $\to$ CH3OH

Given that for this reaction, $\Delta$Hr = -ve

CH4 + 2O2 → CO2 + 2H2O,    $\Delta$H = x ....(1)

CH3OH + ${3 \over 2}$ O2 → CO2 + 2H2O,    $\Delta$H = y ....(2)

Eqn. (1) – Eqn. (2)

CH4 + ${1 \over 2}$ O2 $\to$ CH3OH

$\Delta$Hr = x - y = -ve

$\therefore$ x < y
4
MCQ (Single Correct Answer)

### AIPMT 2001

PbO2  $\to$ PbO; $\Delta$G298 < 0
SnO2 $\to$ SnO;  $\Delta$G298 > 0

Most probable oxidation state of Pb and Sn will be
A
Pb4+, Sn4+
B
Pb4+, Sn2+
C
Pb2+, Sn2+
D
Pb2+, Sn4+

## Explanation

PbO2 → PbO

Pb has +4 oxidation state In PbO2.

Pb has +2 oxidation state In PbO.

Here $\Delta$G is negative that is why reaction is spontaneous and Pb4+ reduces Pb2+ thus, Pb2+ is most probable oxidation state of Pb.

SnO2 $\to$ SnO;

Sn has +4 oxidation state In SnO2.

Sn has +2 oxidation state In SnO.

$\Delta$G is positive that is why reaction is not spontaneous thus, +4 oxidation state of Sn is more stable as it does not change to +2 oxidation state spontaneously.

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