EASY
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IMPORTANT
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Assume all the reactions/process given below are spontaneous at temperature 'T' :-

Reaction I: PCl3g+Cl2gPCl5g

Reaction II : CO2g+CaOsCaCO3s

Reaction III : N2g+H2g2NH3g

The above reactions are Exothermic.

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Important Points to Remember in Chapter -1 - Thermodynamics from Embibe Experts Gamma Question Bank for Engineering Chemistry Solutions

1. Thermodynamics:

The study of energy changes between system and surrounding is called thermodynamics. There are three types of systems, such as open, closed and isolated.

2. Zeroth Law of Thermodynamics:

If two systems are each in thermal equilibrium with a third one, they are also in thermal equilibrium with each other.

3. First Law of Thermodynamics:

Heat supplied Q to a system is equal to algebraic sum of the change in internal energy ΔU of the system and mechanical work W done by the system.

(i) Q=W+ΔU W=PdV; ΔU=nCvΔT

(ii) Sign convention:

(a) Heat absorbed by the system positive

(b) Heat rejected by the system negative

(c) Increase in internal energy (i.e., rise in temperature) positive

(d) Decrease in internal energy (i.e., fall in temperature) negative

(e) Work done by the system positive

(f) Work done on the system negative

(iii) For isochoric process:

(a) In this process, V= constant.

(b)  PT 

(c)  W=0,

(d) Q=ΔU=μCvΔT

(iv) For isobaric process:

(a) In this process, P= constant.

(b) VT

(c) Q=μCpΔT

(d) ΔU=μCvΔT 

(e) W=PV2-V1=μRΔT

(v) For adiabatic process:

(a) In this process, Q=0

(b) PVγ= constant.

(c) TγP1-γ= constant.

(d) TVγ-1= constant.

(e) W=-ΔU=μCVT1-T2=P1V1-P2V2γ-1

(vi) For isothermal process:

(a) In this process, T= constant or ΔT=0

(b)  PV= constant.

(c) ΔU=μCvΔT=0

(d) Q=W=μRTlnV2V1=μRTlnP1P2

(vii) For any general polytropic process:

(a) PVx= constant

(b) Molar heat capacity, C=Cv+R1-x

(c) Work done by gas, W=nRT1-T2x-1=P1V1-P2V2x-1

(d) Slope of P-V diagram (also known as indicator diagram) at any point is dPdV=-xPV

(viii) For cyclic process,

(a) ΔU=0

(b) Q=W

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4. Second Law of Thermodynamics:

(i) Kelvin–Planck statement: 

It is impossible to construct an engine operating in a cycle that will produce no effect other than extracting heat from a reservoir and performing an equivalent amount of work.

(ii) Rudolph Clausius statement:

It is impossible to make heat flow from a body at a lower temperature to a body at a higher temperature without doing external work on the working substance.

5. Heat Engines and Refrigerators:

(i) Carnot engine: It is a hypothetical engine with maximum possible efficiency.

Process 12 and 34 are isothermal.

Process 23 and 41 are adiabatic.

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(ii) Efficiency of a cycle:

(a) For Carnot cycle:

η=Work done by working substanceHeat supplied =WQ1=Q1-Q2Q1=1-Q2Q1=1-T2T1

Diagram Description automatically generated

(b) For refrigerator:

Coefficient of performance, β=Q2W=Q2Q1-Q2=T2T1-T2

Diagram Description automatically generated

6. Bulk Modulus of Gases:

(i) B=ΔP-ΔVV

(ii) Isothermal bulk modulus of elasticity, BIso=-VPVT=constant.

(iii) Adiabatic bulk modulus of elasticity, BAdia=-γVPVBAdia=γBIso.

7. Thermodynamic Processes and Indicator Diagrams:

(i) Work done is least for monoatomic gas (adiabatic process).

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(ii) Air quickly leaking out of a balloon becomes cooler as the leaking air undergoes adiabatic expansion.

(iii) First law of thermodynamics does not forbid the flow of heat from the lower temperature to the higher temperature.

(iv) First law of thermodynamics allows many processes which actually do not happen.