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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