Thermal Physics 3rd Semester previous year Questions with Answers unit 1

Thermal Physics (PHYDSC-202T)

Semester: 3rd Semester

Course: Thermal Physics

Previous year question paper solution unit 1

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UNIT 1

1(a) Why is Cp greater than Cv?

At constant volume, no external work is done. Therefore, all the heat supplied increases the internal energy.

At constant pressure, the gas expands and performs external work. Hence more heat is required to raise the temperature by the same amount.

Cp > Cv

1(b) Limitations of First Law of Thermodynamics

• Does not indicate the direction of heat flow.
• Cannot explain spontaneous processes.
• Does not set the maximum efficiency of heat engines.
• Cannot distinguish reversible and irreversible processes.

1(c) Difference between Isothermal and Adiabatic Process

Isothermal	Adiabatic
Temperature remains constant.	No heat exchange.
Heat is supplied or rejected.	Q = 0
PV = Constant	PVγ = Constant
ΔU = 0 (Ideal Gas)	Temperature changes.

2(a) Relation Between Specific Heats (Cp and Cv)

The First Law of Thermodynamics is

dQ = dU + dW

At Constant Volume:

No work is done because dV = 0.

dQ = dU

dU = Cv dT

At Constant Pressure:

dQ = dU + PdV

Substituting dU = Cv dT

Cp dT = Cv dT + PdV

Using Ideal Gas Equation

PV = RT

PdV = RdT

Therefore

Cp dT = Cv dT + RdT

Cp − Cv = R

Hence, Cp − Cv = R is called Mayer's Relation.

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2(b) Work Done During Isothermal Expansion

An isothermal process is a process in which temperature remains constant.

PV = nRT

Since temperature is constant,

P = nRT / V

The work done is

W = ∫ P dV

Substituting the value of P

W = ∫ (nRT / V) dV

W = nRT ∫ dV / V

After integration,

W = nRT ln(V₂ / V₁)

Also,

W = nRT ln(P₁ / P₂)

Conclusion:

• Temperature remains constant.
• Change in internal energy is zero.
• Heat absorbed equals work done.

Q = W