Thermodynamics

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New answer posted•a year ago

Thermodynamics Class 11th

An ideal gas undergoes a cyclic process whose indicator diagram is as shown in the figure. Find the work done by the gas in 1 complete cycle.

16 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Workdone by gas = area enclosed in P-V diagram
= π (2)² lt-atm

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New answer posted•a year ago

Thermodynamics Class 11th

The speed of sound in a particular gas at a temperature 27°C is 340 m/s. Then, the speed of sound in the same gas at a temperature 90°C is:

6 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

v = √ (γRT/M) ; V? /V? = √ (T? /T? ) ; V? ' = V√ (363/300)

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New answer posted•a year ago

Thermodynamics NEET

Which are thermodynamics processes according to Class 11 Physics Thermodynamics?

7 Views|Posted a year ago

Asked by Pallavi Pathak

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Answered by

Pallavi Pathak | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Thermodynamics NEET

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

8 Views|Posted a year ago

Asked by Pallavi Pathak

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Answered by

Pallavi Pathak | Contributor-Level 10

a year ago

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 significanc

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New answer posted•a year ago

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:

8 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

[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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New question posted•a year ago

Thermodynamics NEET

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

3 Views|Posted a year ago

Asked by Pallavi Pathak

New answer posted•a year ago

Thermodynamics Class 11th

Match List I with List II.

                  List I                                                                                  List II

           (a) Isothermal                                                                 (i) Pressure constant

           (b) Isochoric                                                                    (ii) Temperature constant

           (c) Adiabatic                                                                   (iii) Volume constant

           (d) Isobaric                                                                      (iv) Heat content is constant

           Choose the correct answer from the options given below:

1 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

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:

7 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

ΔH / ΔU = C? / C? = 75 / 25 = 3

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New answer posted•a year ago

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:

8 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

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

5 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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