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.

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

Among the following, intensive property is (properties are)

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 6

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 7

An ideal gas in a thermally insulated vessel at internal pressure $= P_{1},$ volume $= V_{1}$ and absolute temperature $= T_{1}$ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are $P_{2}, V_{2}$ and $T_{2}$ , respectively. For this expansion

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 8

An ideal gas is expanded from

(p_{1}, V_{1}, T_{1})

(p_{2}, V_{2}, T_{2})

under different conditions. The correct statement(s) among the following is (are) :

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 9

Assuming that water vapour is an ideal gas, the internal energy change

(∆ U)

when,

1 mole

of water is vapourised at

1 bar

pressure and

100 ° C

, (Given, molar enthalpy of vapourisation of water at

1 bar

and

373 K = 41 kJ

m o l^{- 1}

and

R = 8 .3 J {mol}^{- 1} K^{- 1}

) will be:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 10

A piston filled with $0.04$ mol of an ideal gas expands reversibly from $50.0 mL$ to $375 mL$ at a constant temperature of $37.0 ° C .$ As it does so, it absorbs $208 J$ of heat. The values of $q$ and $w$ for the process will be:

$(R = 8.314 J {mol}^{- 1} K)$ $(\ln 7.5 = 2.01)$

$(R = 8.314 J {mol}^{- 1} K) (\ln 7.5 = 2.01)$

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 11

△ U

equal to:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-3>Q 14

The value of $\log_{10} K$ for a reaction $A ⇌ B$ is

Given:

\begin{array}{l} ∆_{r} H_{298 K}^{o} = - 54.07 kJ {mol}^{- 1}, ∆_{r} S_{298 K}^{o} = 10 J K^{- 1} {mol}^{- 1}, R = 8.314 J K^{- 1} {mol}^{- 1}, \\ 2.303 \times 8.314 \times 298 = 5705 \end{array}

Among the following, the state function(s) is/are

Important Questions on Thermodynamics

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