State the second law of thermodynamics.

The second law of thermodynamics can be expressed in terms of entropy. It can be stated as “the entropy of an isolated system increases during a spontaneous process”. The total entropy change ( ∆ S total ) for the system and the surroundings of a spontaneous process is given by, ∆ S total = ∆ S system + ∆ S surrounding > 0 When a system is in equilibrium, the entropy is maximum and the change in entropy, ∆ S = 0 . Hence, entropy of a spontaneous process increases till it reaches a maximum and at equilibrium the change in entropy is zero.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25 ° C is

298 × 10 7 × 8 . 31 × 2 . 303 log 2 .

2 . 303 × 298 × 0 . 082 log 2 .

2 . 303 × 298 × 0 . 082 log 0 . 5 .

8 . 31 × 10 7 × 298 × 2 . 303 log 0 . 5 .

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:

with increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

with increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

with increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

HARD

JEE Main/Advance

IMPORTANT

Earn 100

The value of $Δ H_{transition}$ of $C$ (graphite) $⟶ C$ (diamond) is $1.9 kJ / mol$ at $25 ° C$ entropy of graphite is higher than entropy of diamond. This implies that:

50% studentsanswered this correctly

Important Questions on Thermodynamics

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-4>Q 113

Which of the following statement(s) is/are false?

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-4>Q 114

For the reaction

2 {Ag}_{2} O (s) ⟶ 4 Ag (s) + O_{2} (g), ΔH

61.17 kJ {mol}^{- 1}

and

Δ S

132 J K^{- 1} {mol}^{- 1} .

Compute the temperature above, which the given reaction will be spontaneous.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-4>Q 115

Select the correct molar entropy at corresponding temperature using following data:
(i) Heat capacity of solid from $0 K$ to normal melting point $200 K$
$\begin{matrix} C_{P, m} (s) = 0.035 T & J K^{- 1} {mol}^{- 1} \end{matrix}$
(ii) Enthalpy of fusion $= 7.5 KJ {mol}^{- 1} .$
(iii) Enthalpy of vaporisation $= 30 KJ {mol}^{- 1} .$
(iv) Heat capacity of liquid form $200 K$ to normal boiling point $300 K$
$C_{P, m} (ℓ) = 60 + 0.016 T J K^{- 1} {mol}^{- 1}$
(v) Heat capacity of gas from $300 K$ to $600 K$ at $1$ atm
$\begin{matrix} C_{P, m} (g) = 50.0 & J K^{- 1} {mol}^{- 1} \end{matrix}$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-4>Q 116

Select correct statement?

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-4>Q 117

For water, the maximum number of phases that can be in equilibrium is

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 2

At constant volume,

6.0

mol of

H_{2}

gas at

0 ° C

and

100 kPa

was heated to

250 kPa .

The molar heat of

H_{2}

(C_{P}) = 28.9 J

{mol}^{- 1}

(Assume that the heat capacity values do not change with temperature). The final temperature of the

H_{2}

gas and the change in entropy of the process are:

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 3

Hydrazine used in rocket fuels can be obtained by the reaction of ammonia and hydrogen peroxide according to the following equations

$2 {NH}_{3} (g) + H_{2} O_{2} (I) ⟶ N_{2} H_{4} (I) + 2 H_{2} O (I) (Δ H_{reaction} ° = - 241 kJ / mol)$

If $Δ H °$ (formation) of ${NH}_{3}, H_{2} O_{2}$ and $H_{2} O$ are $- 46.1, - 187.8$ and $- 285.8 kJ / mol$ respectively, $Δ H °$ for the decomposition of hydrazine into $N_{2}$ and $H_{2}$ is

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 6

One mole of Ideal gas $(C_{P, m} = 15 {JK}^{- 1} {mole}^{- 1})$ . follow the process as shown in figure. Predict the following:

Question Image

Calculate work done, Heat of process.

The value of ΔHtransition of C (graphite) ⟶C (diamond) is 1.9 kJ/mol at 25 °C entropy of graphite is higher than entropy of diamond. This implies that:

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

The value of $Δ H_{transition}$ of $C$ (graphite) $⟶ C$ (diamond) is $1.9 kJ / mol$ at $25 ° C$ entropy of graphite is higher than entropy of diamond. This implies that: