Hello!!

No,it is not in the syllabus of jee ,we have very little time in hand and it is not okay if you get confused with the syllabus at this phase check out our page and get every syllabus related solution -

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-syllabus/amp

Thanks

We can define thermodynamics, as t he branch of Physics that deals with heat and temperature, and their relation to energy, work, radiation, and properties of matter.

To be specific, it explains how thermal energy is converted to or from other forms of energy and how matter is affected by this process. Thermal energy is the energy that comes from heat. This heat is generated by the movement of tiny particles within an object. The faster these particles move, the more heat is generated.

There are four laws of thermodynamics and are given below:

• Zeroth law of thermodynamics
• First law of thermodynamics
• Second law of thermodynamics
• Third law of thermodynamics

Hello dear student,

Thermodynamics basically deals with the energy. the branch of physical science that deals with the relations between heat and other forms of energy (such as mechanical, electrical, or chemical energy), and, by extension, of the relationships between all forms of energy. . It also says that energy of a system is always conserved.  I hope this helps

Thank you

Suppose we consider an Ideal gas. (Ah, one more definition, Ideal gas is a gas in which we neglect attraction between particles, therefore, no potential energy, thus here internal energy just refers to total kinetic energy) and make it undergo Isothermal process, then BY DEFINITION, temperature or Average kinetic energy of the molecules remains a constant. If Average kinetic energy is a constant, then the total kinetic energy must also remain a constant. So look at the internal energy of this gas. Since internal energy for an ideal gas is just it’s total kinetic energy, by definition, the internal energy must be a constant. Does that make sense now?

However, if we break that assumption, that particles are not attracted to each other, then potential energy comes into the picture. Now, in an isothermal process, even though temperature (average kinetic energy) is a constant, it’s potential energy can definitely change, thus now it’s internal energy can definitely change.

A practical example would be during phase change. Suppose you consider water boiling. Since water is a liquid, you can definitely NOT neglect the attraction between particles and hence potential energy cannot be neglected at all. Hence during boiling, all the heat energy supplied (or taken up by the liquid) increases the potential energy of the system (makes particles farther and farther away), but keeps the total kinetic energy a constant.

So the temperature definitely remains a constant, thus by definition this is an isothermal process, but since potential energy is increasing (like crazy) the internal energy of the system is definitely increasing.

Remember that the first law of thermodynamics states (Q+W+u^1)=(u^2)(Q+W+u^1)=(u^2)

Where u^u^ is internal energy.

By definition of an adiabatic process, the only way that internal energy can change is via work, which is defined as the integral of P dV

Before we can solve this integral, we need to formulate a relationship between pressure and volume under the condition of an adiabatic process. Consider then, the differential form of the first law of thermodynamics (done by considering the limit as the increment of heat and work added to the gas goes to zero, and allowable under the assumption of reversibility), namely

dQ=dW+dU=0=PdV+dUdQ=dW+dU=0=PdV+dU

For an ideal gas, another equation of note is

dU=nCvdTdU=nCvdT

from the definition of Cv=dUdTCv=dUdT

Utilizing the ideal gas equation, we see that

nCvdT=nCvnR(PdV+VdP)nCvdT=nCvnR(PdV+VdP)

Once again substituting into the differential form, we see that

Cv+RRPdV+CvRVdP=0Cv+RRPdV+CvRVdP=0

Which, upon relatively simple integration, results in the equation

PVγ=P0Vγ0PVγ=P0V0γ

and allows integration to find that

P0Vγ0γ−1(V1−γ0−V1−γ1)P0V0γγ−1(V01−γ−V11−γ)

No matter how many miles of helium is mixed with any number of moles of hydrogen there wont formation of any new compounds or molecules. Because helium is a noble gas. Which means it is in the most stable state possible. Hence you would still end up with two moles of helium and n moles of hydrogen.

Dear Aspirant,

Thermodynamics is a very important chapter and you will find a lot of questions from this topic in your school exams and also in various competitive exams like JEE, etc.

Here are some topics you should prepare for:

1. All Laws of thermodynamics
2. Cp-Cv=R
3. Basics of Carnot Engine
4. Isothermal,Isobaric,Isochoric and Adiabatic process
5. Change in Enthalpy and Work done.
6. Refrigerator.

Hope this helps. Thank you.

N2(g) + 3H2(g) ------> 2NH3(g) ; rH° = -92.4 KJ/mol

enthalpy of formation of NH3 means heat released in the formation of 1 mole of NH3 .

so, divide both sides by 2 in above reaction

1/2N2(g) + 3/2H2(g) ---> NH3(g) ;

Therefore , fH° = rH°/2 = -92.4/2 = 46.2 KJ/mol

Thank you.