Here are essential chemistry questions for class 11 with answers for Chapter 6: Chemical Thermodynamics. These key questions are based on the CBSE board syllabus and correspond to the latest Grade 11 Chemistry syllabus. By practicing these key questions for Class 11, students can quickly review all of the ideas covered in this chapter and prepare for the annual class exams 11 as well as other entrance exams such as NEET and JEE.
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Lesson 11 Important Chemical Thermodynamics Questions with Answers
Short answer type questions
1.At 100 °C and 1 bar pressure, 18.0 g of water are completely evaporated, the enthalpy change is 40.79 kJ mol–1. What is the enthalpy change to vaporize two moles of water under the same conditions? What is the standard enthalpy of vaporization of water?
responder
The amount of water is 18.0 g and the pressure is 1 bar.
As we all know, 18.0 g is H2O = 1 Mol H2Ö
Enthalpy change to vaporize 1 mol H2O = 40.79 kJmol–1
∴ Enthalpy change for evaporating 2 moles of H2O = 2 × 40,79 kJ = 81 358 kJ
Standard vaporization enthalpy at 100℃ and 1 bar pressure,
DEvaporatorH2O = + 40.79 kJ mol–1.
2.One mole of acetone requires less heat to vaporize than 1 mole of water. Which of the two liquids has the greater enthalpy of vaporization?
responder
Acetone requires less heat to vaporize due to the weak attraction between the molecules.
As a result, the water has a higher vaporization enthalpy.
It can be represented as:
DEvaporator(water) > ΔEvaporator(acetone)
3.Molar standard enthalpy of formation, ΔFHӨ is just a special case of the reaction enthalpy ΔRHÖ. is that dRHӨ for the following reaction is equal to ΔFHº? Justify your answer. CaO(s) + CO2(g) → CaCO3(S); DFHӨ = –178,3 kJ mol–1
responder
No, the ΔRHӨ for the given reaction is not the same as ΔFSIM.
The standard enthalpy change for the formation of one mole of a compound from its elements in its most stable states (reference states) is called the standard molar enthalpy of formation ΔRSIM.
CaO(s) + C(s) + 3/2 O2(g) → CaCO3(S)
This reaction is different from the given reaction. Hence ΔRH ≠ DFSIM.
4.The value of ΔFIS a NH3es - 91.8 KJ mol–1. Calculate the enthalpy change for the following reaction:
2NH3(g) → Norden2(g) +3H2(Gramm)
responder
DFH Ө is given for training. For the reverse reaction, ΔFHӨ changes sign because the opposite of an exothermic reaction is an endothermic reaction.
So, for a mole, ΔFHӨ for the decomposition is -(-91.8) = 91.8.
However, two moles decompose here.
Daher DFHӨ = 2 × 91,8 = 183,6 KJ mol–1.
5.Enthalpy is a large property. In general, when the enthalpy of an overall reaction A→B along a path is ΔrH and ΔrH1, ΔrH2, ΔrH3….. represent enthalpies of intermediate reactions leading to product B. What will be the relationship between ΔrH for the overall reaction and ΔrH?1, ΔrH2….. etc. for intermediate reactions?
responder
In general, when the enthalpy of an overall reaction A→B along a path is ΔrH and ΔrH1, ΔrH2, ΔrH3…..represent reaction enthalpies that lead to the same product B via a different route, then ΔrH = ΔrH applies1+ΔrH2+ΔrH3+…..
For a general reaction, Hess' law of constant sum of heat can be represented as
6.The enthalpy of atomization for the CH reaction4(g)→ C(g) + 4H (g) é 1665 kJ mol–1. What is the binding energy of the C-H bond?
responder
Atomization enthalpy of 4 mol C-H bonds = 1665 kJ mol–1
∴ C−H bond energy, per mol = 1665 kJ mol–1/4 = 416.2 kJmol–1.
7.Use the following data to calculate ΔgridHÖfor NaBr.
DsubHÖfor metallic sodium = 108.4 kJ mol–1
Natriumionisationsenthalpie = 496 kJ mol–1
Bromine enthalpy of electron gain = – 325 kJ mol–1
Bromine bond dissociation enthalpy = 192 kJ mol–1
DFHÖfor NaBr(s) = -360.1 kJ mol–1
responder
According to Hess's law,
DfHÖ=DsubHÖ+ ΔIEHÖ+DPlatoÖ+DtgHÖ+ of
DsubHÖfor In metal = 108.4 kJ/mol
that is, Na = 496 k/mol
DtgHÖthe Br = – 325 kJ mol–1
DPlatoÖthe Br = 192 kJ mol–1
DfHÖfor NaBr = -360.1 kJ mol–1
DfHÖ=DsubHÖ+ IE of Na + ΔPlatoÖ+DtgHÖ+ of
-360,1 = 108,4 +496 + 96 + (-325) – U
U = +735,5 kJ/mol
8.Since ΔH = 0 for the mixture of two gases. Explain whether the diffusion of these gases among themselves in a closed container is a spontaneous process or not.
responder
It's a spontaneous process. Although the enthalpy change is zero, the change in randomness or disorder (i.e. ΔS) increases. As a result, in the equation ΔG = ΔH - TΔS, the TΔS term is negative. As a result, ΔG will be negative.
9.Heat has a random effect on a system, and temperature is the measure of the average chaotic motion of the particles in the system. Write down the mathematical relationship that relates these three parameters.
responder
Heat has a random effect on a system, and temperature is the measure of the average chaotic motion of the particles in the system.
The mathematical relationship relating these three parameters is ΔS = qRdo/T
Here ΔS = entropy change
QRdo= reversible heat of reaction
T = Temperature
10The enthalpy increase of the environment is equal to the enthalpy decrease of the system. Will the temperatures of the system and the environment be the same in thermal equilibrium?
responder
Yes, when the system and the environment are in thermal equilibrium, their temperatures are the same.
11At 298 K. Kp for the reaction of N2Ö4(g) ⇄ 2NO2(g) is 0.98. Predict whether the reaction will be spontaneous or not.
responder
For a spontaneous reaction, ΔrG° is -ve.
ΔrG° = – RT ln KPage= – RT In (0,98)
In (0.98) is - 0.02
∴ΔrG° = –RT × –0,02
∴ Δ G° will be positive.
Therefore, the reaction will not be spontaneous.
12A 1.0 mol sample of a monatomic ideal gas undergoes a cyclic expansion and compression process as shown in the figure. 6.1. What is the value of ΔH for the entire cycle?
responder
The change in internal energy during a cycle is zero.
ΔU = 0
When the system returns to its initial state, it says no work has been done.
The enthalpy for a cyclic steady-state process is a single value at each stage, but varies at different stages.
ΔH = 0
13The standard molar entropy of H2The (l) is 70 JK–1mol–1. The standard molar entropy of H2OR(s) greater than or less than 70 JK–1mol–1?
responder
A fixed form of H2Or is it ice cream. In ice, the molecules are less random than in liquid water.
Then the molar entropy of H2O(s) < molar entropy of H2Or I). The standard molar entropy of
H2Or(s) is less than 70 JK–1mol–1.
14Identify state functions and path functions among the following: enthalpy, entropy, heat, temperature, work, free energy.
responder
State functions: enthalpy, entropy, temperature, free energy
Path functions: heat, work
15.The molar enthalpy of vaporization of acetone is smaller than that of water. Because?
responder
Acetone has no hydrogen bonding, so the intermolecular forces are weaker, causing it to boil or vaporize quickly, lowering the molar enthalpy of vaporization. Because acetone does not have a polar O-H bond, it also has low enthalpy. Water has a non-polar region and strong hydrogen bonding.
sixteen.How much of ΔRG e DRGRAMMÖwill be zero in equilibrium?
responder
DRG will always be zero.
DRGRAMMÖis zero for K = 1 because ΔRGRAMMÖ= – RT lnK
DRGRAMMÖwill be non-zero for other values of K.
17Predict the change in internal energy for an isolated system at constant volume.
responder
There is no energy transfer as heat or work in an isolated system,
then w=0 and q=0.
According to the first law of thermodynamics
ΔU = q + w = 0 + 0 = 0
ΔU = 0
18Although heat is a function of displacement, the heat absorbed by the system under certain specified conditions is independent of displacement. What are these conditions? Explain.
responder
(i) A constant volume
According to the first law of thermodynamics: q = ΔU + (–w)
(–w) = pΔV
∴q = ΔU + pΔV
ΔV = 0 since the volume is constant.
∴qv= ΔU + 0 ⇒ qV = ΔU = change in internal energy
(ii) At constant pressure
QPage= ΔU + pΔV
Mas, ΔU + pΔV = ΔH
∴ qp = ΔH = enthalpy change.
Therefore, at constant volume and constant pressure, the heat change is a state function because it is equal to the internal energy change and enthalpy change, respectively, which are state functions.
19The expansion of a gas in a vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 liter of ideal gas expands isothermally in vacuum until the total volume is 5 liters.
responder
(–w) = pagrama(V2– v.1) = 0 × (5 –1) = 0
For q = 0 isothermal expansion
According to the first law of thermodynamics
q = ΔU + (–w)
0 = U + 0
then ΔU = 0
20heat capacity (CPage) is an extensive property, but specific heat (c) is an intensive property. How will the relationship between CPageand c for 1 mol of water?
responder
For water, heat capacity = 18 × specific heat
o cPage= 18 × c
Specific heat, c = 4.18 Jg–1k–1
heat capacity CPage= 18 × 4,18 JK–1= 75,3 JK–1.
21The difference between Cp and Cv can be derived from the empirical relationship H = U + pV. Calculate the difference between Cp and Cv for ten moles of an ideal gas.
responder
For 1 mole of gas
CPage- Cv= R
For n moles of gas, the relationship is CPage- Cv= nR = 10 × 4184 J
CPage- Cv= 41,84 J.
22If burning 1 g of graphite produces 20.7 kJ of heat, what is the molar enthalpy change? Clear character meaning too.
responder
Molar enthalpy change of graphite = enthalpy change for 1 g carbon ×
molar mass of carbon
ΔH = – 20,7 kJ g–1× 12 g mol–1
ΔH = – 2,48 × 102kJ had–1.
The exothermicity of the reaction is indicated by the negative sign of ΔH.
23The net enthalpy change of a reaction is the amount of energy required to break all bonds in the reactant molecules minus the amount of energy required to form all bonds in the product molecules. What is the enthalpy change for the following reaction?
H2(g) + br2(g) → 2HBr(g)
Since the binding energy of H2, Brother2and HBr is 435 kJ mol–1, 192 kJ had–1and 368 kJ times–1or.
responder
DRHÖ= binding energy of H2+ binding energy Br2- 2 × HBr binding energy
DRHÖ= 435 + 192 – (2 × 368) kJ mal–1
DRHÖ= –109 kJ mol–1.
24The vaporization enthalpy of CCl4is 30.5 kJ mol–1. Calculate the heat required to vaporize 284 g of CCl4at constant pressure. (Molar mass of CCl4= 154 gmol–1).
responder
QPage= ΔH = 30,5 kJ mol–1
∴ Heat required to vaporize 284 g of CCl4
25The reaction enthalpy for the reaction:
2 hours2(g) +O2(g) → 2H2O(l)éΔrHº= – 572 kJ mal–1.
What is the standard enthalpy of formation of H2Oh me)?
responder
The enthalpy change for the next reaction is the standard enthalpy of formation of H2Or according to the definition of the standard enthalpy of formation.
or the standard enthalpy of formation of H2The (l) will be half the enthalpy of the given equation, ie ΔrHΘ will also be halved.
26How much work is done on an ideal gas confined in a bottle when compressed by a constant external pressure, pramain one step as shown in Fig. 6.2. Explain graphically.
responder
The work done on an ideal gas can be calculated from the p-V diagram shown in the figure. 6.6. The work done corresponds to the hatched area ABVUEvII.
27How do you calculate the work done on an ideal gas in compression when the pressure change is in infinite steps?
responder
A process or change is said to be reversible if it is carried out in such a way that the process can be reversed at any time by an infinitesimal change.
If the pressure is not constant and the changes occur in infinite steps (reversible conditions) during the initial volumetric compression, then VUEfor the last volume, vF, the work done can be calculated using a pV plot. The shaded area represents the work done on the gas.
28Graph the change in potential energy/enthalpy in the following processes.
(a) Throwing a stone from the floor to the ceiling.
(b) ½ H2(g) + ½ Cl2(g) ⇄ HCl(g); ΔrHº= –92.32 kJmol–1
In what process does the change in potential energy/enthalpy contribute to spontaneity?
responder
(a) When we throw a stone from the floor to the ceiling, we have to add energy to the stone.
(b) If heat is generated in a reaction, this indicates that the process is followed by a decrease in energy.
The energy partially increases in (a) while it decreases in (b). As a result, in process (b), the enthalpy change contributes to the spontaneity.
29The enthalpy diagram for a specific reaction is shown in the figure. 6.3. Is it possible to determine the spontaneity of a reaction from the given diagram? Explain.
responder
No, enthalpy is not the only criterion for spontaneity; we must also consider entropy.
301.0 mole of a monatomic ideal gas expands from state (1) to state (2) as shown in the figure. 6.4. Calculate the work done in expanding the gas from state (1) to state (2) at 298 K.
responder
Since the process takes place in an infinite number of steps, it can be seen from the diagram that the change is reversible.
\(\begin{matrix}{l}w=-2.303RT log \frac{V_{2}}{V_{1}}\end{matrix} \)
\(\begin{matrix}{l}\frac{V_{2}}{V_{1}}=\frac{p_{1}}{p_{2}}=\frac{2}{1}=2 \end{matrix} \)
∴
\(\begin{matriz}{l}w=-2.303RT log\frac{p_{1}}{p_{2}}\end{matriz} \)
\(\begin{matriz}{l}w=-2.303\times 8.314Jmol^{-1}\times 298K^{-1}\times log2\end{matriz} \)
= -1717,46 J
31An ideal gas can expand from 10 L to 50 L in one step against a constant pressure of 2 bar. Calculate the work done by the gas. If the same dilation were performed reversibly, would the work done be more or less than in the previous case? (Because 1 L bar = 100 J).
responder
The work done can be calculated as
w = –Pex(VF– v.UE)
w = – 2 × 40 = – 80 bar = – 8 kJ
The minus sign indicates that the system is working in the area. Since the internal and external pressures are almost the same in each stage, the work done in the reversible expansion will be larger.
Long answer questions
1.Derive the relationship between ΔH and ΔU for an ideal gas. Explain each term in the equation.
responder
When heated, the volume of solids and liquids does not change significantly. Therefore, when the volume changes, ∆V is negligible.
∆H = ∆U + P∆V
∆H = ∆U + P(0)
∆H = ∆U
When gases are involved in the reaction, the difference between the change in internal energy and the change in enthalpy becomes significant.
Imagine a chemical reaction that takes place at constant temperature T and constant pressure P. Let's assume the volume of the reactants is VAand the number of moles in the reactants is nA. Likewise, the volume of the product V isB, and the number of moles in the product is nB.
From the ideal gas equation we know:
pV = nRT
pVA= northART
pVB= northBRT
For this reason,
pVA– PVB= northATR-nBRT
p(VA– v.B)=RT(nA– NorteB)
p∆V = ∆nGrammRT
∆H = ∆U + p∆V
∆H = ∆U + ∆nGrammRT
2. Extensive properties depend on the amount of substance, but intensive properties do not. Explain whether the following properties are extensive or intensive.
Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.
responder
- Mass is an extensive property because it depends on the amount of matter.
- Internal energy is proportional to volume, so it's an extensive property.
- Since the pressure is independent of the amount of substance, it is an intensive property.
- Heat capacity is an extensive property while molar heat capacity is an intensive property since it is the ratio of two extensive properties.
- Density is an intensive property because it is the ratio of two extensive properties.
- Because the mole fraction is the ratio of two extensive properties, it is an intensive property.
- Specific heat is a powerful property because it is independent of the amount of substance.
- An intensive property is a relationship of two extensive properties, so molarity is an intensive property.
- Temperature is also an intensive property since it is independent of the amount of substance.
3.The lattice enthalpy of an ionic compound is the enthalpy when one mole of a gaseous ionic compound dissociates into its ions. It is impossible to determine it directly through experimentation. Propose and explain an indirect method to measure the lattice enthalpy of NaCl(s).
responder
When one mole of an ionic compound dissociates into its ions, the enthalpy change associated with this process is called the lattice enthalpy.
Von+Kl–(s) → Through+(g) + Cl–(Gramm); ∆gridHº= + 788 hJmol–1
The lattice enthalpy for NaCl can be calculated as follows:
I. Sublimation of Metallic Sodium
Na (s) → Na (g), ∆subHº= 108.4 kJ mal–1
ii. Ionization of sodium atoms
Na(g) → Na+(g) + mi–(Gramm); ∆THº= 496 kJ mal–1
iii. Dissociation of the chlorine molecule, the reaction is half the enthalpy of bond dissociation.
½ cl2(g) → Cl(g); ½ ∆ConnectionHº= 121 kJmol–1
4. At (g) + e–(g) → Kl–(grams); Electron gained from chlorine atoms. The enthalpy of electron gain, ∆For exampleHΘ = – 348,6 kJ mol–1.
4.ΔG is the net energy available to do useful work and is therefore a measure of "free energy". Prove mathematically that ΔG is a measure of the free energy. Find the unit of ΔG. If a reaction has a positive enthalpy change and a positive entropy change, under what conditions will the reaction be spontaneous?
responder
∆SIn total= ∆SSystem+ ∆Sthe sum
∆SIn total= ∆SSystem+ ( – ∆HSystem/T)
T∆SIn total= T∆SSystem– ∆HSystem
For spontaneous change, ∆SIn total> 0
∴ T∆SSystem– ∆HSystem> 0
-(∆HSystem– T∆SSystem) > 0
But ∆HSystem– T∆SSystem= ∆GSystem
∴ –∆GSystem> 0
∆GSystem= ∆HSystem– T∆SSystem< 0
∆HSystem= enthalpy change of a reaction
T∆SSystem= Energy that is not available to perform useful work
∆GSystem= Energy available for useful work
- If ∆G is negative (< 0), the process is spontaneous.
- If ∆G is positive (> 0), the process is not spontaneous.
The unit of ∆G is Joule.
The reaction will be spontaneous at high temperature.
5.Show graphically the total work done in an expansion as the state of an ideal gas changes reversibly and isothermally from (pi,Vi) to (pf,Vf). Using a pV diagram, compare the work done in the above case with the work done against a constant external pressure pf.
responder
(i) The reversible work is represented by the combined areas
mi
(ii) Work against constant pressure, pf is represented by the area
work (i) > work (ii)
FAQs
What are the important topics of thermodynamics class 11 chemistry? ›
- Thermodynamic terms.
- Applications.
- Measurement of change in U and H: Calorimetry.
- Enthalpy change.
- Enthalpies for different types of reactions.
- Spontaneity.
- Gibbs energy change & Equilibrium.
The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. This means that heat energy cannot be created or destroyed.
What is thermodynamics chapter of chemistry class 11? ›The laws of Thermodynamics Class 11 deals with energy changes in macroscopic level involving a large number of molecules. It is important to note that this branch of science is not concerned with the rate of energy transformation carried out instead deals with the initial and final stages of a system.
What is the name of CH 6 in Class 11 chemistry? ›NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics.
Is thermodynamics hard class 11? ›Besides Ionic, Electrochemistry and Thermodynamics are also considered tough.
Is thermodynamics a difficult chapter? ›There is only a minor difference, in Physics, thermodynamics study is combined with Heat in the form of a chapter. It is a very difficult chapter and needs a good amount of practice.
Is thermodynamics chemistry hard? ›In some cases, thermodynamics is hard because the concepts are hard and students often have numerous misconceptions. Many students think an isothermal process is a process without heat transfer. Some concepts cannot be jettisoned from the class in order to make it easier.
What are the first 3 laws of thermodynamics? ›1st Law of Thermodynamics - Energy cannot be created or destroyed. 2nd Law of Thermodynamics - For a spontaneous process, the entropy of the universe increases. 3rd Law of Thermodynamics - A perfect crystal at zero Kelvin has zero entropy.
What is the value of R? ›The gas constant value is given by R = 8.3144598(48) J⋅mol^−1⋅K^−1.
Is thermodynamics Chapter easy? ›It is not a very tough chapter and requires a lot of practice. The theory is relatively very easy to understand and once you see solved examples you would be able to sail through the numerical portion easily. The chapter been given good weightage in JEE Advanced and will fetch you marks easily.
What is thermodynamics class 11 easy explanation? ›
What is Thermodynamics? The branch which deals with the movement of energy from one form to the other and the relation between heat and temperature with energy and work done is called as thermodynamics.
What is the main topic of thermodynamics? ›Thermodynamics is the study of the relations between heat, work, temperature, and energy. The laws of thermodynamics describe how the energy in a system changes and whether the system can perform useful work on its surroundings.
Which is the most important chapter in chemistry class 11? ›...
Weightage of Class 11 Chemistry in JEE Mains
- Chemical Bonding.
- Atomic Structure.
- Periodic Properties.
- General Organic Chemistry.
- Hydrocarbons.
Be diligent in studying, practice all the numerical problems and equations. Pay extra attention to Organic Chemistry and study all the reactions thoroughly. Go through the NCERT book and concentrate on all topics. Practise all the exercises and questions.
What is gas equation class 11? ›The ideal gas equation is formulated as: PV = nRT. In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the total amount of ideal gas that is measured in terms of moles, R is the universal gas constant, and T is the temperature.
Is class 11th easy? ›But the difficulty level of class 11th will dawn upon you sooner or later. 11th is very different in terms of the course content as compared to what you have studied till 10th. Difficulty level is higher if you have chosen the science stream as now it is no more general science.
Is Class 11 hard or 12? ›Hi 11th standard is comparatively tougher than 12th. so before starting 11th class go through the basics and formulas related to 11th class and try to solve questions of math, physics, chemistry from NCERT book.
Is class 11 science very hard? ›Science (PCM or PCB)
Science stream has a reputation of most difficult stream in class 11th and 12th. Students learn Science till 10th standard, so they are aware of the basics of this subject.
- Physical Chemistry is the hardest branch of chemistry.
- It involves the study of physical properties and constitution of matter, the laws of chemical combination, and theories governing chemical reactions.
In Chemistry I found the Equilibrium chapter the hardest for NEET!
Which chapter is easy in chemistry? ›
To secure good marks in CBSE 12th Board Chemistry, students can cover easiest chapters first that include Biomolecules, Solutions,Chemistry in Everyday LIfe and Polymers. If you rate chapter on Biomolecules, it can be ranked lowest in difficulty level.
How to pass thermodynamics exam? ›Studying to Pass Thermodynamics Exams
Review your textbook, practice example problems, homework problems, and review solutions over and over. This is all you need to do to be able to pass Thermodynamics class.
Chemistry is a challenging subject for most people, but it doesn't have to be. The number one reason people struggle with chemistry is that they don't approach it the right way. Below we'll explore proven strategies and techniques that will, if applied, improve your ability to study and learn chemistry.
Are chemistry exams hard? ›AP Chemistry consistently ranks among the most difficult AP courses. The course covers many topics, while prioritizing essential scientific practices. AP Chemistry test data overlooks factors like the quality of teachers and schools. The AP Chemistry exam comprises both multiple-choice and written sections.
Are there 3 or 4 laws of thermodynamics? ›There are four laws of thermodynamics. They talk about temperature, heat, work, and entropy. They are used in thermodynamics and other sciences, for example chemistry.
What are the 3 main systems of thermodynamics? ›There are three types of systems in thermodynamics: open, closed, and isolated.
What is the law of entropy? ›The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases.
What is R formula for? ›The R method is most often used to find the extrema (maximum and minimum) of combinations of trigonometric functions, since the extrema of a basic trigonometric function are easy to work with (both sine and cosine have a minimum of -1 and a maximum of 1).
What is the value of E? ›The exponential constant is an important mathematical constant and is given the symbol e. Its value is approximately 2.718. It has been found that this value occurs so frequently when mathematics is used to model physical and economic phenomena that it is convenient to write simply e.
What is the value of F? ›The F value is used in analysis of variance (ANOVA). It is calculated by dividing two mean squares. This calculation determines the ratio of explained variance to unexplained variance. The F distribution is a theoretical distribution.
What are the basic questions in thermodynamics? ›
- State Zeroth Law of Thermodynamics. ...
- How will you Classify Thermodynamic System? ...
- What is the Influence of Heat Losses or Gains on the Surface of Conducting Body? ...
- Explain the Stages of Combustion in SI Engines.
For PHYSICS : The easiest chapters among all the syllabus is the modern Physics, Radioactivity, kinematics( it is very scoring), Thermodynamics, waves and optics.
Can I study thermodynamics one day? ›You can complete in one day for your semester exams but for that you should have the basic knowledge so that you don't have to go thoroughly the book. You just need to see the topics in syllabus, match it in book and take a small reading so that you can get the central idea of topic.
What is the formula of thermodynamics 11? ›Thermodynamics formula
C=q/dT; q = Heat absorbed, dT = rise in temperature.
Thermodynamic Process
Adiabatic Process – A process where no heat transfer into or out of the system occurs. Isochoric Process – A process where no change in volume occurs and the system does no work.
Thus, in the equation ΔU=q+w w=0 and ΔU=q. The internal energy is equal to the heat of the system. The surrounding heat increases, so the heat of the system decreases because heat is not created nor destroyed.
What are the four types of thermodynamics? ›- Isothermal process.
- Isobaric process.
- Isochoric process.
- Adiabatic process.
There are several types of thermodynamic processes, including (a) isothermal, where the system's temperature is constant; (b) adiabatic, where no heat is exchanged by the system; (c) isobaric, where the system's pressure is constant; and (d) isochoric, where the system's volume is constant.
What are the 1 and 2 laws of thermodynamics? ›Two fundamental concepts govern energy as it relates to living organisms: the First Law of Thermodynamics states that total energy in a closed system is neither lost nor gained — it is only transformed. The Second Law of Thermodynamics states that entropy constantly increases in a closed system.
How to pass class 11 chemistry exam? ›- Allocate your study time wisely.
- Develop a study plan.
- Order the topics starting with the highest level of difficulty.
- Must Read. ...
- Solve sample papers by simulating an exam environment.
- Take proper sleep, stay hydrated and eat nutritious food.
How to score full marks in class 11? ›
Solve Sample Papers
By solving sample papers, students can get good practice for the final exam. Try to solve previous year's question papers also. It will help to get an idea of the question pattern and marking scheme.
- Chemical Thermodynamics.
- Equilibrium.
- Redox Reactions.
- Hydrogen.
- P-Block.
- Organic Chemistry.
- Hydrocarbons.
- Environmental Chemistry.
Don't keep too many tasks for a day. Each stream in class 11 has 5 subjects, so make sure you divide them equally in a week and spend equal time accordingly. Don't take too much pressure, stop picking dozen of topics in a go but instead divide them in 5-6 topics and study wisely.
How do you get 180 in chemistry? ›Answer: Nearly 15 questions are asked from all three, including physical, organic, and inorganic chemistry. Almost all questions asked are from NCERT class 11th and 12th. Answer: Take out at least two hours for revision every day. Do not procrastinate.
What are the 2 equations for gas laws? ›Charles's law: VT = constant at constant P and n. Avogadro's law: Vn = constant at constant P and T.
How many gas laws are there in class 11? ›The five gas laws are: Boyle's Law, which provides a relationship between the pressure and the volume of a gas. Charles's Law, which provides a relationship between the volume occupied by a gas and the absolute temperature.
What are all the gas laws Class 11? ›The study of the behavior of gases has led to some important generalisations called the Gas Laws. The four main laws are Boyle's Law, Charles' Law, Gay Lussac's Law and Avogadro's Law. Boyle's Law: Boyle's Law expresses the relationship between volume and pressure.
What is the most important topics for thermodynamics? ›Important Topics of Thermodynamics
First Law of Thermodynamics. Second Law of Thermodynamics. Third Law of Thermodynamics. Enthalpy and Enthalpy Change.
- Basic Concepts of Thermodynamics.
- Thermodynamics and Energy.
- Dimensions and Units.
- Closed and Open Systems.
- Energy.
- CLOSED. SYSTEM. m = const. mass. energy.
- CONTROL. VOLUME.
- Properties of a System.
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To enhance your preparation, you can focus on these specific chapters from Class 11 Chemistry:
- Chemical Bonding.
- Atomic Structure.
- Periodic Properties.
- General Organic Chemistry.
- Hydrocarbons.
What topics are covered in thermodynamics? ›
What is thermodynamics? Thermodynamics is the study of the relations between heat, work, temperature, and energy. The laws of thermodynamics describe how the energy in a system changes and whether the system can perform useful work on its surroundings.
What are the 4 laws of thermodynamics called? ›There are four laws of thermodynamics. They talk about temperature, heat, work, and entropy.
What are the 3 thermodynamic processes? ›In Thermodynamics, types of processes include: Isobaric process in which the pressure (P) is kept constant (ΔP =0). Isochoric process in which the volume (V) is kept constant (ΔV =0). Isothermal process in which the temperature (T) is kept constant (ΔT =0).
What is the hardest chapter in chemistry class 11? ›- Physical Chemistry is the hardest branch of chemistry.
- It involves the study of physical properties and constitution of matter, the laws of chemical combination, and theories governing chemical reactions.
Be diligent in studying, practice all the numerical problems and equations. Pay extra attention to Organic Chemistry and study all the reactions thoroughly. Go through the NCERT book and concentrate on all topics. Practise all the exercises and questions.