The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

\\({Q}_{1}={T}_{0}{S}_{0}+\cfrac{1}{2}{T}_{0}{S}_{0}=\cfrac{3}{2}{T}_{0}{S}_{0}\\)\\({Q}_{2}={T}_{0}(2{S}_{0}-{S}_{0})={T}_{0}{S}_{0}\\) and \\({Q}_{3}=0\\)\\(\eta=\cfrac{W}{{Q}_{1}}=\cfrac{{Q}_{1}-{Q}_{2}}{{Q}_{1}}\\)\\(=1-\cfrac{{Q}_{2}}{{Q}_{1}}=1-\cfrac { { T }_{ 0 }{ S }_{ 0 } }{ \cfrac { 3 }{ 2 } { T }_{ 0 }{ S }_{ 0 } } =\cfrac { 1 }{ 3 } \\)

5 mole of an ideal gas expands reversibly from a volume of \\(8d\\) to \\(80d\\) at a temperature of \\(27\\). The change in entropy is:-(A) \\(41.57{ JK }^{ -1 }\\)(B) \\(-95.73{ JK }^{ -1 }\\)(C) \\(95.73{ JK }^{ -1 }\\)(D) \\(-41.57{ JK }^{ -1 }\\)

The correct option is CWe have,\\(n=5,v_1=8dm^3, v_2=80dm^3, T=300k\\)ConstantThus the change in entropy is:\\( \Delta s=nR \\) in \\(\dfrac{v_2}{v_1}\\)\\(=5\times8.314\times \dfrac{80}{8}\\)\\(=95.73jk^{-1}\\)

Define entropy. Give its units.

Entropy is a measure of randomness or disorder of the system.The greater the randomness, the higher the entropy.It is state function and extensive property.Its unit is \\(JK^{-1}mol^{-1}\\).

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

From Kelvin Plank statement and Claussius's statement we can say that heat can not be flow form lower temperature to higher temperature until and unless we are adding any external source like heat pump, which justifies \\({SECOND \ LAW \ OF \ THERMODYNAMICS}\\)

Calculate the change in entropy when \\(1\ g\\) of ice at \\(0^oC\\) is heated to form water at \\(40^oC\\)

The change in entropy would be given as \\(\Delta S=\Delta S_1+\Delta S_2\\) \\(\Delta S_1\\) is to melt the ice and \\(\Delta S_2\\) is to rise the temperature. Thus we get \\(\Delta S=\displaystyle\dfrac{mL}{T}+m\int^{T_2}_{T_1}\dfrac{d T}{T}\\)\\(\Delta S=\displaystyle\dfrac{1\times 80}{273}+1\times\\)\\(\ln \dfrac{313}{273}=0.28+0.13=0.41\ cal/^oC\\)

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

Heat is a form of energy that gets transferred from one medium or object to another medium/object. Suppose if we a colder body and a hotter body and if it brought in contact, the heat gets transferred from the hotter body to colder body until the system of hotter & colder attains equilibrium.When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then The heat lost by the hotter body will be equal to the heat gained by the colder body.option D is correct

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

The second law of Thermodynamic states that the total entropy of an isolated system increases over time, or it remains same in ideal cases (where the system underdoes a reversible process or when it is on a steady state). The first law of Thermodynamic tells the basic definition of internal energy and states the law of conservation of energy. The second law of Thermodynamic deals with the direction of natural process. it states that a natural process is not reversible. For e.g., heat flows hot to cold region, and reverse, unless some external work is done the system.In Terms of EntropyIn a reversible process, a very small increase in the entropy \\((dS)\\) of a system is defined to result from a small heat transfer to a closed divided by the common temperature \\((T\\) of the system and surrounding that supply of the heat:\\(dS=\dfrac {dQ}{T}\\) (closed system)Various Statements of the Planck's Law:1. Statement of Kelvin: According to this statements it is not possible to make an engine in which the working substance can take heat from the source and change it completely into, work and remain unchanged itself in one complete cycle. It is concluded from this statement that it is not possible to get work continuously from heat source. Actually the sink in necessary and it necessary to release. Some heat to the sink. This statements is based on the principle of carnot heat engine.2. Statement of Clausius: According to this statements, it is an impossible process that in cyclic process the working substance transfer heat from low temperature object to high temperature object directly in the absence of external work.It is concluded from this statement that transfer of heat from a low temperature object high temperature object on its own is not possible. For this it is important that external work be done on the working substance. This statements is based on the principle of refrigerator.Equivalence of Clausius and Kelvin Statements:Let there be an engine violating the Kelvin statement: i.e., on that drains heat and converts it completely into work in a cyclic manner without any other results. Now, pairing it with a reverse Carnot engine.The newly created engine (consists of two engines) transfers heat \\(\Delta Q=Q\left(\dfrac {1}{\eta}-1\right)\\) from the colder region (sink) to the hotter region (source), which violates the statements of Clausius. Thus, violation of Kelvin statement means violation of Clausius statement, i.e., Clausius statement implies the Kelvin statement. Similarly, we can prove that the Kelvin Statement implies the Clausius statement and hence both the statements are equivalent.

Entropy is a measure of

Entropy is a measure of disorder and randomness

Variation of heat of reaction with temperature is given by Kirchhoff's equation.Which of the following is not the equation for kirchhoff's ?

Kirchoff's law describes the variation of enthalpy of a reaction with the change in temperature. And at constant pressure, if the heat capacities \\((c_p)\\) do not vary with temperature then the change in enthalpy is a function of the difference in temperature and the heat capacities.\\(\Delta{H_2}=\Delta{H_1}+\Delta{c_p}(T_2-T_1)\\)So, equations \\(C\\) and \\(D\\) are the form of the above equation.The equation \\(B) \ dH=TdS+VdP\\) is derived from the second law of thermodynamics.

Other statements of Second Law of Thermodynamics | Definition, Examples, Diagrams

definition

Second Law in terms of efficiency

The second Law of thermodynamics gives a fundamental limitation to the efficiency of a heat engine and the co-efficient of performance of a refrigerator. In simple terms, it says that the efficiency of a heat engine can never be unity.

definition

Kelvin Planck statement

No process is possible whose sole result is the absorption of heat into a reservoir and its complete conversion into work.

definition

Clausius statement

No process is possible whose sole result is the transfer of heat from a colder body to a hotter body.

definition

Qualitative statement of Second Law of thermodynamics

Qualitative statement of Second Law of Thermodynamics:
It is not possible to design a heat engine which works in cyclic process and whose only result is to take heat from a body at a single temperature and convert it completely into mechanical work.

example

Physical examples of Second Law of Thermodynamics

Physical examples of Second law of thermodynamics are as follows:
1. An internal combustion engine in a car converts chemical energy from the gasoline to heat to forward motion. The best efficiency we could get is something on the order of 35%, and that's the absolute best we can do, ignoring friction losses and the like. (The exact maximum efficiency is dependent on the temperature of the combustion and the heat sink, i.e. ambient).
2. Heat flows from bodies with higher temperature to the lower temperature and not vice versa.
3. Work can be converted completely into heat as in case of a resistor but heat cannot be completely converted into work as is in the case of a heat engine.
4. A drop of dye when placed in a beaker of water will eventually result in an evenly coloured solution, even without stirring. The dye molecules diffuse and spread itself as evenly as possible throughout the volume of water.

definition

Identify whether a process follows or violates Second Law of Thermodynamics

The process which follow second law of thermodynamics:
1. A drop of dye when placed in a beaker of water will eventually result in an evenly coloured solution, even without stirring. The dye molecules diffuse and spread itself as evenly as possible throughout the volume of water.
2. Making snowballs will always make ones hands colder, because thermal energy always flows in a hot-to-cold direction. No material can spontaneously gain thermal energy from colder surroundings. This is because it would require the energy to become spontaneously more concentrated in order for the opposite process to occur. That is, this would involve a spontaneous increase in organization of molecules or energy which would mean the opposite of energy spreading out.
A device that violates the Second Law of Thermodynamics is formally known as a perpetual motion machine of the second kind. A perpetual motion machine is a hypothetical machine that can do work indefinitely without an energy source. This kind of machine is impossible, as it would violate the first or second law of thermodynamics.

Other statements of Second Law of Thermodynamics

definition

Second Law in terms of efficiency

definition

Kelvin Planck statement

definition

Clausius statement

definition

Qualitative statement of Second Law of thermodynamics

example

Physical examples of Second Law of Thermodynamics

definition

Identify whether a process follows or violates Second Law of Thermodynamics

Quick Summary With Stories

Second Law of Thermodynamics

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

5 mole of an ideal gas expands reversibly from a volume of $8 d$ to $80 d$ at a temperature of $27$ . The change in entropy is:-
(A) $41.57 J K^{- 1}$
(B) $- 95.73 J K^{- 1}$
(C) $95.73 J K^{- 1}$
(D) $- 41.57 J K^{- 1}$

Define entropy. Give its units.

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

Calculate the change in entropy when $1 g$ of ice at $0^{o} C$ is heated to form water at $4 0^{o} C$

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

If the enthalpy change for the transition of liquid water steam is 30 kJ mo $l^{- 1}$ at $27^{\circ} C,$ the entropy change for the process would be :

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

Entropy is a measure of

Variation of heat of reaction with temperature is given by Kirchhoff's equation.Which of the following is not the equation for kirchhoff's ?

definition

Second Law in terms of efficiency

definition

Kelvin Planck statement

definition

Clausius statement

definition

Qualitative statement of Second Law of thermodynamics

example

Physical examples of Second Law of Thermodynamics

definition

Identify whether a process follows or violates Second Law of Thermodynamics

Quick Summary With Stories

Second Law of Thermodynamics

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

5 mole of an ideal gas expands reversibly from a volume of 8d to 80d at a temperature of 27. The change in entropy is:- (A) 41.57JK−1 (B) −95.73JK−1 (C) 95.73JK−1 (D) −41.57JK−1

Define entropy. Give its units.

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

Calculate the change in entropy when 1 g of ice at 0oC is heated to form water at 40oC

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

If the enthalpy change for the transition of liquid water steam is 30 kJ mol−1 at 27∘C, the entropy change for the process would be :

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

Entropy is a measure of

Variation of heat of reaction with temperature is given by Kirchhoff's equation.Which of the following is not the equation for kirchhoff's ?

5 mole of an ideal gas expands reversibly from a volume of $8 d$ to $80 d$ at a temperature of $27$ . The change in entropy is:-
(A) $41.57 J K^{- 1}$
(B) $- 95.73 J K^{- 1}$
(C) $95.73 J K^{- 1}$
(D) $- 41.57 J K^{- 1}$

Calculate the change in entropy when $1 g$ of ice at $0^{o} C$ is heated to form water at $4 0^{o} C$

If the enthalpy change for the transition of liquid water steam is 30 kJ mo $l^{- 1}$ at $27^{\circ} C,$ the entropy change for the process would be :