In the videos you decribrd tempature to actually be just he kinetic motion of the particles but,when the fusion takes place [at the periods where the teparature remains constant], we learnt that the energy is used to overcome the intermolecular bonds so when these bonds are being overcome sholdn't the kinetic energy also increase in turn causing the temparutare to also increse.

Open in App

Solution

During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules

Suggest Corrections

Similar questions

Q. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?

Increae in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?

(a) Increase (b) No effect

Q. Assertion :Viscosity of liquids decreases as the temperature rises. Reason: At high temperature, molecules have high kinetic energy and can overcome the intermolecular forces to flow faster.

Consider the D−T reaction (deuterium−tritium fusion)

(a) Calculate the energy released in MeV in this reaction from the data:

= 2.014102 u

= 3.016049 u

(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Q. Consider the DT reaction (deuteriumtritium fusion).

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e + n

(a) Calculate the energy released in MeV in this reaction from the data:

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmans constant, T = absolute temperature.)

Join BYJU'S Learning Program