# Entropy Change in Irreversible Process

## Trending Questions

**Q.**Following table represents critical temperature of some gases. Arrange these gases in their increasing order of liquefaction.

Gas | H2 | He | N2 | O2 |

Tc (K) | 33.2 | 5.3 | 126 | 154.3 |

- He<N2<H2<O2
- H2<He<N2<O2
- He<H2<N2<O2
- O2<N2<H2<He

**Q.**For irreversible expansion of an ideal gas under isothermal condition, the correct option is :

- Δ∪≠0, ΔS total=0
- Δ∪=0, ΔS total=0
- Δ∪≠0, ΔStotal≠0
- Δ∪=0, ΔS total≠0

**Q.**Why is spontaneous process irreversible

**Q.**The incorrect option for free expansion of ideal gas under isothermal condition is

- ΔH=0
- ΔS=0
- ΔT=0
- w=0

**Q.**The total entropy of a system and its surrounding increases, if the process is

- reversible
- irreversible
- exothermic
- endothermic

**Q.**Assertion:

Critical temperature of CO2 is 304 K, it cannot be liquified above 304 K.

Reason:

At a certain temperature, Volume∝1Pressure

- Both assertion and reason are true and reason is the correct explanation of assertion.
- Both assertion and reason are true but reason is not the correct explanation of assertion.
- Assertion is true but reason is false.
- Both assertion and reason are false.

**Q.**

1 mole of a triatomic non-linear ideal gas is subjected to an adiabatic process from initial temperature 400 K and pressure 32 atm to final pressure 2 atm. If the expansion is carried out against vaccum, the entropy change (in cal K−1) for the system is [ln2=0.7]

**Q.**What is the cell entropy change of the following cell ?

Pt(s)|H2(g, 1atm)|CH3COOH, HCl||KCl(aq)|Hg2Cl2(s)|Hg

EMF of the cell is found to be 0.0045 V at 298 K and temperature coefficient is 3.4×10−4 V K−1

- 105 kJ K−1
- 205.7 J K−1
- 5.45 J K−1
- 65.61 J K−1

**Q.**Two moles of an ideal gas are expanded irreversibly and isothermally at 37∘ C until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

ΔStotal(system + surrounding) is (R=253 J/molK, ln 2=0.7, ln 3 = 1.1):

- - 0.667 J/K
- 0.667 J/K
- 22.7 J/K
- \N

**Q.**Given that:

S∘H2=131 JK−1mol−1S∘Cl2=223 JK−1mol−1and S∘HCl=187 JK−1mol−1

The standard entropy change in formation of 1 mole of HCl(g) from H2(g) and Cl2(g) will be:

(Given, reaction for formation of HCl : H2+Cl2→2HCl)

- 187 JK−1
- 374 JK−1
- 20 JK−1
- 10 JK−1

**Q.**Can anyone say what is reversible inhibition and irreversible inhibition?

**Q.**Two moles of an ideal gas are expanded irreversibly and isothermally at 37∘ C until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

ΔStotal(system + surrounding) is (R=253 J/molK, ln 2=0.7, ln 3 = 1.1):

- - 0.667 J/K
- 0.667 J/K
- 22.7 J/K
- \N

**Q.**Starting with the same initial conditions, one mole of an ideal monoatomic gas expands reversibly from volume V1 to V2 in two different ways. The work done by gas is w1, if the process is purely isothermal, w2 if purely isobaric and w3 if purely adiabatic, then:

- w2>w1>w3
- w2>w3>w1
- w1>w2>w3
- w1>w3>w2

**Q.**The correct folrmula of Q by using Born-Haber cycle for the preparation of MgBr2(s) will be:

S = heat of sublimation of Mg(s)

I = ionisation energy of Mg(g)

D = dissociation energy of Br2(g)

-E = electron affinity of Br(g)

-U = lattice energy of MgBr2(s)

Q = heat of formation of MgBr2(s)

- Q=S+I+D−2E−U
- Q=S−I+D−2E−U
- Q=S+I+D−E−2U
- None of these

**Q.**

A diatomic ideal gas initially at 272 K is given 100 cal heat due to which system did 209 J work. Molar heat capacity (Cm) of gas for the process is

5/2 R

5R

5/4 R

3/2 R

**Q.**For a chemical reaction A+B⇌C+D (ΔrHo=80 kJ mol1) the entropy change ΔrSo depends on the temperature T (in K) as ΔrSo=2T (JK−1 mol−1). Minimum temperature at which it will become spontaneous is ________ K .

**Q.**

Two moles of ideal monoatomic gas was taken through isochoric heating from 100 K to 800 K. If the process is carried out irreversibly (one step) then ΔSsystem + ΔSsurrounding is

[Given : ln 2 = 0.7 and R = 2 cal/mol K]

0

- 3.675 cal/K
- 7.35 cal/K
- None of these

**Q.**For a spontaneous process, which of the following is true?

- ΔSsys is positive
- ΔSsurr is positive
- ΔStotal is positive
- ΔStotal is negative

**Q.**When 1 mol of an ideal gas is compressed to half of its volume, its temperature becomes double; then the change in entropy (ΔS) would be:

- CVln2
- CPln2
- (CV−R)ln2
- CVRln2

**Q.**

Identify the correct statement regarding a spontaneous process.

- Lowering of energy in the process is the only criterion for spontaneity.
- For a spontaneous process in an isolated system, the change in entropy is positive.
- Exothermic process are always spontaneous.
- Endothermic process are never spontaneous.

**Q.**Two mole of an ideal gas is expanded irreversibly and isothermally at 37∘C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. ΔStotal (system + surrounding) is :

- 0.52J/K
- 0
- 22.52J/K
- none of the above

**Q.**Two mole of an ideal gas is expanded irreversibly and isothermally at 37∘C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. ΔStotal (system + surrounding) is :

- −0.52 J/K
- 0.52 J/K
- 22.52 J/K
- 0

**Q.**1 mole of an ideal gas is allowed to expand isothermally at 27oC until its volume is tripled. Calculate ΔSsys and ΔSuniv under the following conditions:

(a) the expansion is carried out reversibly.

(b) the expansion is a free expansion

**Q.**

Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change (ΔSsystem + ΔSsurrounding) for given process. [Given: ℓn2 = 0.70 amd R=8.0 J/mol/K]

56 J/K

14 J/K

16 J/K

None of these

**Q.**Which of the following statements is correct?

- (∂H∂T)p−(∂U∂T)v=R
- (∂H∂T)p>(∂U∂T)v
- (∂U∂V)T for ideal gas is zero
- All of the above

**Q.**

For a spontaneous process.

- ΔGsytem= +ve only
- ΔGsytem= zero
- ΔStotal= - ve
- ΔStotal= +ve

**Q.**Select correct option regarding change in entropy of system (ΔSsys) and of the surroundings (ΔSsurr).

- ΔSsys>0, ΔSsurr<0
- ΔSsys<0, ΔSsurr>0
- ΔSsys=0, ΔSsurr=0
- ΔSsys>0, ΔSsurr=0

**Q.**Entropy change of a system is determined by the initial and final states only, irrespective of how the system has changed its states.

If true enter 1 else 0

- 1

**Q.**Two moles of an ideal gas are expanded irreversibly and isothermally at 37∘ C until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

ΔStotal(system + surrounding) is (R=253 J/molK, ln 2=0.7, ln 3 = 1.1):

- - 0.667 J/K
- 0.667 J/K
- 22.7 J/K
- 0

**Q.**

Two moles of ideal monoatomic gas was taken through isochoric heating from 100 K to 800 K. If the process is carried out irreversibly (one step) then ΔSsystem + ΔSsurrounding is

[Given : ln 2 = 0.7 and R = 2 cal/mol K]

0

3.675 cal/K

7.35 cal/K

None of these