Explanation:

Closed System: There is no mass transfer across the system boundary but energy transfer takes place into or out of the system.

Open System: In this type of system both mass and energy transfer takes place across the boundary of the system.

Isolated System: In this type of system neither mass nor energy crosses the boundary of the system.

Explanation:

A control volume is a fixed, identifiable region in space through which fluid flows. The boundary of the control volume is called a control surface.

Types of System

Control Mass System (Closed System)

• It is a system of fixed mass with a fixed identity
• This type of system is usually referred to as "closed system"
• There is no mass transfer across the system boundary
• Energy transfer may take place into or out of the system

Control Volume System (Open System)

• It is a system of fixed volume
• This type of system is usually referred to as "open system” or a "control volume"
• Mass transfer can take place across a control volume
• Energy transfer may also occur into or out of the system
• A control volume can be seen as a fixed region across which mass and energy transfers are studied
• Control Surface- It is the boundary of a control volume across which the transfer of both mass and energy takes place
• The mass of a control volume (open system) may or may not be fixed

Explanation:

Charles’ Law states that for an ideal gas under constant pressure, the volume of the gas is directly proportional to its temperature.

PV = mRT

Now for constant Pressure, if we plot a graph between V v/s T then it is a straight line

Charles plotted the graph experimentally for different pressures and found out the following graph

PV = mRT

When T = 0 K, Volume is zero.

From the above graph we can see that at T= –273.15 ˚C, Volume become zero.

We know that 0 K is –273.15 ˚C which is nothing but the Absolute zero temperature.

Absolute temperature is temperature measured using the Kelvin scale where zero is absolute zero The zero point is the temperature at which particles of matter have their minimum motion and can become no colder (minimum energy).

Concept:

There are two thermodynamics functions:

Path function:

• Heat and work are path functions.
• Their magnitude depends on the path followed during a process as well as end states.

Point function:

• Pressure, temperature, volume and internal energy etc. are point function.
• They depend on the end states only, not on the path followed.

Calculation:

Given:

Here in all the three processes, end states are fixed i.e. A and B only.

So, during all the three processes i.e. A-1-B, A-2-B, and A-3-B, change in internal energy will be the same.

ΔU₁ = ΔU₂ = ΔU₃

According to first law of thermodynamics:

δQ = δW + ΔU

Work done = Area under P-V curve

Area₁ > Area₂ > Area₃

W₁ > W₂ > W₃

Concept:

In this type of scaling problems,

Equate gradient of both the scales between Boiling point and freezing point

\({\left( {\frac{{Boiling\;point - x}}{{Boiling\;point - freezing\;point}}} \right)_{scale1}} = {\left( {\frac{{Boiling\;point - x}}{{Boiling\;point - freezing\;point}}} \right)_{scale2}}\)

Calculation:

Let's consider a new scale as scale 1 and the Celcius scale as scale 2.

Given:

The boiling point on new scale = 400°C

The freezing point on new scale = 100°C

To find the temperature, at which both the Celsius and the new temperature scale reading would be the same

Now,

Equating the gradient of both the scales between Boiling point and freezing point

\(\begin{array}{l} \left( {\frac{{400 - x}}{{400 - \left( { 100} \right)}}} \right) = \left( {\frac{{100 - x}}{{100 - 0}}} \right)\\ x = \;-50^\circ \;C \end{array}\)

∴ The reading at which both the Celsius and the new temperature scale reading would be the same will be -50°C

Explanation:

Work transfer and heat transfer are a path function.

Example:

It is possible to take a system from state 1 to state 2 along many quasi-static paths such as A, B or C. Since the area under each curve represents the work for each process, the amount of work involved in each case is not a function of the end states of the process and it depends on the path which the system follows in going from state 1 to state 2. For this reason, work is called a path function, and δW is an inexact or imperfect differential.

According to the question:

For the given problem statement i.e. “work done in an adiabatic process”, the path is fixed i.e. adiabatic process.

And for a given path work done will be dependent on the end states only.

So, the work done in an adiabatic process between two given end states depends on end states only.

or, from first law of thermodynamics,

δQ = ΔU + δW, 

for an adiabatic process δQ = 0.

i.e. 0 = ΔU + δW

⇒ δW = - ΔU = U1 - U2

Since, work done is equal to change in internal energy and internal energy is a point function or state function.

So, the work done in an adiabatic process between two given end states depends on end states only.

Explanation:

Intensive Property: 

• These are the properties of the system which are independent of mass under consideration.
• For e.g. Pressure, Temperature, density

Extensive Properties:

• The properties that depend on the mass of the system under consideration.
• For e.g Internal Energy, Enthalpy, Volume, Entropy

Thermodynamic work:

The non-flow work done by the system (only on the quasi-static process) is given by:

\({W} = \mathop \smallint \limits_1^2 pdV\)

Here Pressure P is the intensive property and ΔV is changed in volume i.e. change in the extensive property.

So, thermodynamic work is the product of an intensive property and change in an extensive property.

Explanation:

Polytropic Process is represented by PVⁿ = C

• n = 0 ⇒ P = C ⇒ Constant Pressure Process (Isobaric Process)
• n = 1 ⇒ PV = C ⇒ Constant Temperature Process (Isothermal process)
• n = γ ⇒ PV^γ = C ⇒ Adiabatic Process
• n = ∞ ⇒ V = C ⇒ Constant Volume Process (Isochoric process)

Area under P-V diagrams represents the work done.

For compression between two volumes, we can see that area is maximum for the adiabatic process.

So work done on the gas will be maximum for the adiabatic process.

Concept:

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant. Energy can be transformed from one form to another but cannot be created or destroyed.

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings

δQ = ΔU + δW

ΔU = δQ - δW

According to first Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

ΣQ = ΣW

Net work done = Net heat in the cycle

Calculation:

Given, heat transfer in the cyclic process is +20 kJ, -5 kJ, -10 kJ and +15 kJ

Net work done in the cycle = Net heat in the cycle

W_net = +20 kJ - 5 kJ - 10 kJ + 15 kJ = +20 kJ

Concept:

According to the Ideal gas equation for a polytropic process:

\({P_1}V_1^n = {P_2}V_2^n\)

\( \Rightarrow {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^n } = \frac{{{P_2}}}{{{P_1}}}\)

Taking log on both sides

\(n \ln \left( {\frac{{{V_1}}}{{{V_2}}}} \right) = \ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

\( \Rightarrow n = \frac{{\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)}}{{\ln \left( {\frac{{{V_1}}}{{{V_2}}}} \right)}}\)

Work done for Polytropic process is, \({{W}_{1-2}}=\frac{{{P}_{1}}{{V}_{1}}\;-\;{{P}_{2}}{{V}_{2}}}{n\;-\;1}\)

Heat transfer for Polytropic process is, \(Q = W\left( {\frac{{\gamma\;- \;n}}{{\gamma\;-\;1}}} \right)\)