Thermodynamics

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3 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Due to weak force of attraction between molecules, acetone requires less heat to vaporise. Hence, water has higher enthalpy of vaporization.

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3 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Enthalpy of a reaction is the energy change per mole for the process.

18 g of H2O = 1 mole ΔHvap = 40.79 kJ/ mol

 Enthalpy change for vapourising 2 moles of H2O = 2 x 40.79 = 81.58 kJ ΔH°vap = 40.79 kJ mol-1.

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3 months ago

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alok kumar singh

Contributor-Level 10

(i) Reversible work is represented by the combined areas

(ii) Work against constant pressure, Pis represented by the area

          

Work (i) > Work (ii). 

The Approach While Dealing With the Concept of Thermodynamics

Since the concept of Thermodynamics and the terminologies of Chemistry are a bit new to the students, they should first learn the names of different chemical components and how to write the chemical equations properly through NCERT Exemplars and Solutions books.

While writing the chemical equations, they might make some mistakes. To avoid making errors, they shoul

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

We know,

ΔStotal = ΔSsys + ΔSsurr

When a system is in thermal equilibrium with its surroundings, the surroundings' temperature is the same as the system's. Furthermore, a rise in the enthalpy of the surroundings equals a decrease in the system's enthalpy. As a result of the entropy shift in the environment,

ΔSsurr = ΔHsurr/T = -ΔHsys/T

ΔStotal = ΔSsys = (-ΔHsys/T)

Rearranging the above equation:

ΔStotal = TΔSsys - ΔHsys

For spontaneous process,

ΔStotal > 0, so 

TΔSsys - ΔHsys> 0

⇒ ( - ΔHsys - TΔSsys) > 0

The above equation can be written as

- ΔG > 0

ΔG

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

A contrast is drawn in thermodynamics between extensive and intense qualities. An extensive property is one whose value is proportional to the amount or size of matter in the system. Extensive properties include mass, volume, internal energy, enthalpy, and heat capacity, to name a few.

Properties that are independent of the amount or size of matter present are known.

As though they were intensive properties Temperature, density, and pressure, for example, are intense properties. A molar property? m, is the value of an extensive property of the system for 1 mol of the sub

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4 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q1= W+Q2

W=Q1-Q2>0

Q1>Q2>0

We can also write Q21

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4 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Change in internal energy for process A to B

dU=nCvdT=nCv (dT)=nCv (TB-TA)

work done from A to B  = area under the PV curve which is maximum for path I

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) For isothermal dT= 0 so T=constant

For an ideal gas dU = change in internal energy = nCvdT=0

From first law of thermodynamics dQ= dU+dW

dQ= dW

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