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Thermodynamics NEET

Which are thermodynamics processes according to Class 11 Physics Thermodynamics?

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Pallavi Pathak

Contributor-Level 10

There are three main processes Isothermal, adiabatic and cyclic process. In isothermal, the system is thermally conductive and the temperature remains constant. In adiabatic process, the system is thermally isolated and there is no change in heat temperature. The system returns to its initial stage in the cyclic process.

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Thermodynamics NEET

What is weightage of Class 11 Physics Chapter 11 Thermodynamics in NEET and JEE Main exams?

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Pallavi Pathak

Contributor-Level 10

In NEET entrance test, the weightage of Class 11 Physics Chapter 11 Thermodynamics is around 7%. Students should thoroughly understand the thermodynamic laws and internal energy. The medical entrance test includes theory and numerically-based questions. For JEE Main, the chapter has more significance, and it has nearly 15% weightage of the total Physics syllabus.

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Thermodynamics Class 11th

n mole of perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A->B : Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and Pressure changes from P₁ to P₂.

B->C : Isobaric compression at pressure P₂ to initial volume V₁.

C->A : Isobaric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is:

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alok kumar singh

Contributor-Level 10

$W_{A B} = n R T l n \frac{2 V_{1}}{V_{1}} = n R T l n 2 .$

$W_{B C} = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{1}{2} n R T$

[At B, 2P₂ V₁ = nRT]

$W_{C A} = 0$ [CA is isochoric process].

$W_{A B C A} = W_{A B} + W_{B C} + W_{C A} = n R T (l n (2) - \frac{1}{2})$

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Thermodynamics NEET

What is weightage of Class 11 Physics Chapter 11 Thermodynamics in NEET and JEE Main exams?

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Thermodynamics Class 11th

Match List I with List II.

List I &nb

...more

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alok kumar singh

Contributor-Level 10

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant.

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Thermodynamics Class 11th

Certain fixed quantity of an ideal gas expands by adiabatic manner such that w_gas is equal to -25 kJ and ∆H_gas for the expansion is -75kJ. The ratio of C_p/C_v for this gas will be equal to:

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Vishal Baghel

Contributor-Level 10

ΔH / ΔU = C? / C? = 75 / 25 = 3

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Thermodynamics Class 11th

Given that ∆c H° (combustion enthalpy) for Diamond is -834.8 kJ mole and ∆c H° for graphite is -832.8 kJ/mole, the value of ∆f H°(Diamond) i.e. [C(graphite) → C(diamond)] [in kJ/mole] is:

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Vishal Baghel

Contributor-Level 10

C (graphite) → C (diamond)
ΔH? = Δ? H° (graphite) - Δ? H° (Diamond) = -832.8 - [-834.8] = 2kJ/mole

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Thermodynamics Class 11th

Find the enthalpy of combustion of propane gas from the following data. Bond energy (kJ/mol)
BE_C-H = a₁
BE_O=O = a₂
BE_C=O = a₃
BE_O-H = a₄
BE_C-C = a₅

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Vishal Baghel

Contributor-Level 10

C? H? + 5O? → 3CO? + 4H? O
ΔH = − {6 (C = O) + 8 (OH)} + 2 (C – C) + 8 (C – H) + 5 (O = O)
= 8a? + 5a? + 2a? - 6a? - 8a?

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Thermodynamics Class 11th

The term - TΔS_total at constant temperature and pressure is known as

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Vishal Baghel

Contributor-Level 10

ΔG = -TΔS_universe

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Thermodynamics Class 11th

If the molar specific heat at constant pressure for a polyatomic non-linear gas is x and the molar specific heat at constant volume for a diatomic gas is y, find the value of x y.

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Vishal Baghel

Contributor-Level 10

x = 8R/2, y = 5R/2
x/y = 1.60

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