NEET Chemistry Thermodynamics Questions Solved


Work done in reversible adiabatic process is given by:

(a) 2.303RTlog(V2/V1)                      (b)nR (γ-1)(T2-T1)

(c) 2.303RTlog(V2/V1)                      (d) none of these

Concept Videos :-

#3 | Internal Energy & 1st Law of Thermodynamics

Concept Questions :-

Internal Energy and work done

(b) This is the derived formula for Wrev in adiabatic process.

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Change in enthalpy for reaction,

2H2O2(l) 2H2O(l) + O2(g)

if the heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively is           

(a) -196 kJ/mol              (b) + 196 kJ/mol

(c) +948 kJ/mol             (d) -948 kJ/mol

Concept Videos :-

#4 | Heat
#26 | Entropy
#29 | Entropy Change Calculation: Case II, III & IV

Concept Questions :-

Enthalpy and It's Type

(a) 2H2O2(l) 2H2O(l) + O2(g) H=?

  

         = [(2x-286) + (0) -(2 x -188)]

         =[-572 + 376] = -196 kJ/mol

 

          

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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K)(4.0 atm, 5.0 L, 245 K) with a change in internal energy, U = 30.0 L atm. The change in enthalpy (H) of the process in L atm is:

(a) 40.0

(b) 42.3

(c) 44.0

(d) not defined, because pressure is not constant

Concept Videos :-

#2 | State & Path Functions
#5 | Work
#9 | Work done in Isothermal Reversible Expansion
#10 | Work done in Isothermal Irreversible Expansion
#36 | Solved Problems on Thermodynamics: Set 1
#37 | Solved Problems on Thermodynamics: Set 2

Concept Questions :-

First Law of Thermodynamics

(c) H=U + PV

      ... H2-H1 = U2-U1+(P2V2 - P1V1)

      ...     H = 30 + (4x5 - 2x3) = 44 L atm

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At absolute zero, the entropy of a perfect crystal is zero. This is........... of thermodynamics.

(a) first law               (b) second law

(c) third law              (d) none of these

Concept Videos :-

#27 | Entropy Change Estimation
#28 | Entropy Change Calculation : Case I
#31 | Condition for Spontaneity
#34 | Solved Problems
#40 | Solved Problems on Spontaneity

Concept Questions :-

Spontaneity and Entropy

(c) This is definition of third law of thermodynamics.

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At constant pressure and temperature, the direction of any chemical reaction is one where, the...... decrease.

(a) entropy                (b) enthalpy

(c) Gibbs energy        (d) none of these

Concept Videos :-

#32 | Gibb's Free Energy
#35 | Relation between ΔGo & Keq : II

Concept Questions :-

Gibbs Energy Change

(c) Spontaneous process shows a decrease in G

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The work done by a mass less piston in causing an expansion V (at constant temperature), when the opposing pressure, P is variable, is given by:

(a) W= -PV                (b) W=0

(c) W= -PV                      (d) none of these

Concept Videos :-

#3 | Internal Energy & 1st Law of Thermodynamics

Concept Questions :-

Internal Energy and work done

(a) Wrev-PdV or -PV; note that opposing pressure is not constant throughout.

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Entropy decreases during:

(a) crystallization of sucrose from solution

(b) rusting of iron

(c) melting of ice

(d) vaporization of camphor

(a) During solidification disorder decreases.

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Entropy change of vaporisation at constant pressure is given by:

1. S(v)= HvT                 2. S(v)=UvT3. S(v)= HvT                 4. none of these

Concept Videos :-

#27 | Entropy Change Estimation
#28 | Entropy Change Calculation : Case I
#31 | Condition for Spontaneity
#34 | Solved Problems
#40 | Solved Problems on Spontaneity

Concept Questions :-

Spontaneity and Entropy

(1) This is derived formula.

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The entropy change for the reaction given below, 

2H2 (g) + O2 (g) 2H2O(l)

is ... at 300 K. Standard entropies of H2 (g), O2(g) and H2O(l) are 126.6, 201.20 and 68.0 JK-1mol-1 respectively.

(a) -318.4 JK-1mol-1                           (b) 318.4 JK-1mol-1 

(c) 31.84 JK-1mol-1                             (d) none of these

Concept Videos :-

#27 | Entropy Change Estimation
#28 | Entropy Change Calculation : Case I
#31 | Condition for Spontaneity
#34 | Solved Problems
#40 | Solved Problems on Spontaneity

Concept Questions :-

Spontaneity and Entropy

(a) Sreaction = ΣSproduct-ΣSreactant               = 2 x SH2O-[2xSH2+SO2]

                  = 2 x 68-[2 x 126.6 + 201.20]

                  = -318.4 JK-1mol-1

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If S° for H2, Cl2 and HCl are 0.13, 0.22 and 0.19 kJ K-1mol-1 respectively. The total change in standard entropy for the reaction, H2 + Cl2 2HCl is:

(a) 30 J K-1mol-1                         (b) 40 J K-1mol-1

(c) 60 J K-1mol-1                         (d) 20 J K-1mol-1

Concept Videos :-

#27 | Entropy Change Estimation
#28 | Entropy Change Calculation : Case I
#31 | Condition for Spontaneity
#34 | Solved Problems
#40 | Solved Problems on Spontaneity

Concept Questions :-

Spontaneity and Entropy

(a) S=SP-SR

          =(2 x0.19) - 0.13-0.22

          = 0.03 kJ K-1mol-1

          = 30 JK-1mol-1

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