UNIT 2 | unit 2 gas laws gas processes

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UNIT 2

GAS LAW AND GAS PROCESSES

Question 1) A gas exerts a pressure of 4 kPa on the walls of container 1.When container 1 is emptied into a 12-liter container, the pressure exerted by the gas increases to 10 kPa. Find the volume of container 1. Assume that the temperature and quantity of the gas remain constant.

Answer:

Given,

Initial pressure, P1 = 4kPa

Final pressure, P2 = 10kPa

Final volume, V2 = 12L

According to Boyle’s law, V1 = (P2V2)/P1

V1 = (10 kPa * 12 L)/4 kPa = 30 L

Therefore, the volume of container 1 is 30 L.

Question 2) a gas occupies a volume of 500cm3 at 0°C and 780 mm Hg. What volume (in liters) will it occupy at 80°C and 780 mm Hg?

Answer: Given, V1= 500 cm³

                                               V2 =?

                                                T1= 0°C= 0+273 = 273 K

                                                T2= 90°C= 90+273 = 363 K

Here the pressure is constant and only the temperature is changed.

Using Charles Law,

(V1/ T1)= (V2 / T2)

500 / 273=V2 / 363

V2=500 * 363 / 273

V2= 664.83 cm3

1 cubic centimetre = 0.001 litre =1 x 10-3 litre

∴ 664.83 cubic centimetre = 664.83 x 10-3 = 0.664 litres

Question 3) A tyre containing 20 moles of air and occupying a volume of 60L loses half its volume due to a puncture. Considering that the pressure and temperature remain constant, what would be the amount of air in the deflated tyre?

Answer: Given,

The initial amount of air (n1) = 20 mol

The initial volume of the tyre (V1) = 60 L

The final volume of the tyre (V2) = 30 L

According to Avogadro’s law, the final amount of air in the tyre (n2) = (V2n1)/V1

= 30L * 20 moles / 60 L

The deflated tyre would contain 10 moles of air.

Question 4) Write the applications of the first law of thermodynamics to study the flow process?

Answer: Application of first law to study flow process

A steady ﬂow process is one in which matter and energy ﬂow steadily in and out of an open system. In a steady ﬂow process, the properties of the ﬂow remain unchanged with time, that is, the properties are frozen in time. But, the properties need not be the same in all points of the ﬂow. It is very common for a beginner to confuse the term steady with the term equilibrium. But, they are not the same. When a system is at a steady-state, the properties at any point in the system are steady in time but may vary from one point to another point. The temperature at the inlet, for example, may diﬀer from that at the outlet. But, each temperature, whatever its value, remains constant in time in a steady ﬂow process. When a system is at an equilibrium state, the properties are steady in time and uniform in space. By properties being uniform in space, we mean that a property, such as pressure, has the same value at every point in the system.

A steady ﬂow is one that remains unchanged with time, and therefore a steady ﬂow has the following characteristics:

i. No property at any given location within the system boundary changes with time. That also means, during an entire steady ﬂow process, the total volume Vs of the system remains a constant, the total mass ms. Of the system remains a constant, and that the total energy content Es of the system remains a constant.

Ii. Since the system remains unchanged with time during a steady ﬂow process, the system boundary also remains the same

Iii. No property at an inlet or at an exit to the open system changes with time. That means that during a steady ﬂow process, the mass ﬂow rate, the energy ﬂow rate, pressure, temperature, speciﬁc (or molar) volume, speciﬁc (or molar) internal energy, speciﬁc (or molar)enthalpy, and the velocity of ﬂow at an inlet or at an exit remain constant.

Iv. Rates at which heat and work are transferred across the boundary of the system remain unchanged.

The application of the first law of thermodynamics to steady-flow processes is defined by two fundamental principles:

(a) Steady mass flow rate m = pAu (mass continuity) where p = fluid density

A = duct area

u = fluid velocity

And m = steady mass flow rate (dm/dt)

(This is usually recast as mv = uA since v = 1/p.)

(b) Conservation of energy (the steady-flow energy equation or SFEE)

Q 1-2 –W 1-2 =m [(h 2 -h 1) + ½ (u 2 2 –u 1 2) + g(Z 2 -Z l )]

Where Z = height above datum.

Note that very often gz is negligible compared with other terms and the reduced SFEE is written as

Q 2 -W 2 =m [(h2 –h1) + ½ (u 2 2 –u 1 2 ) ]

In per unit mass flow rate.

Q/m=q, W/m= w

q 2 - w 2 = (h 2 – h 1) + ½ (u 2 2 –u 1 2 )

An even further reduction by negligible kinetic energy terms here

q2 - w2 = (h2 – h1)

Many processes, in reality, approximate closely to a steady flow, for example, steady

Conditions in a steam power plant after the start-up transient is over and

Steady conditions exist at all points in the system.

Here two essential characteristics of steady flow:

(a) m is constant,

(b) The properties at any station are invariable with time.

The one obvious application of the SFEE where gz is prominent is in a hydroelectric plant where a change in z is essential for power production.

Question 5) Explain the difference between isothermal and adiabatic processes?

Answer: The Difference Between Isothermal and Adiabatic Process

Isothermal process	Adiabatic process
An isothermal process is defined as one of the thermodynamic processes which occur at a constant temperature	An adiabatic process is defined as one of the thermodynamic processes which occur without any heat transfer between the system and the surrounding
Work done is due to the change in the net heat content in the system	Work done is due to the change in its internal energy
The temperature cannot be varied	The temperature can be varied
There is a transfer of heat	There is no transfer of heat

Question 6) writes refrigerant properties?

Answer: Refrigerant properties:

Following are the refrigerant properties shown below

1) Low boiling point and Low freezing point.

2) Low specific heat and High latent heat.

3) High critical pressure and temperature

4) Low specific volume to reduce the size of the compressor.

5) High thermal conductivity to compact evaporator and condenser.

6) Non-flammable, non-explosive, non-toxic, and non-corrosive.

7) High miscibility with lubricating oil

8) High COP in the working temperature range.

9) Compatible with a legal requirement

10) Availability and cost

Question 7) Write the Difference between split and packaged units?

Answer: Difference between split and packaged unit

The main difference between a packaged unit system and a split system air conditioner lies in its construction. The packaged unit contains all its components placed together in a single unit. This means all the parts including the thermostat, condenser, compressor, and evaporator are present in the same metallic cabinet.

The split unit is more of a space saver and easy to install as it can be placed anywhere in your basement or simply on the other side of the wall. On the other hand, you need a proper roof location for the placement of the packaged unit in case of central air-conditioning.

Packaged air conditioning units consume more power in comparison to the split unit.

Most of the brands are now offering Inverter technology in their split systems, making the air conditioning units more energy-efficient and super quiet. The inverter splits ACs also have the capability of handling voltage or current fluctuations, unlike the packaged units.

The air handler of the split systems is installed indoors to enhance the life-span of the AC filters and evaporator coils.

Question 8) Explain the adiabatic process formula?

Answer: Adiabatic process derivation

The adiabatic process can be derived from the first law of thermodynamics relating to the change in internal energy dU to the work dW done by the system and the heat dQ added to it.

DU =dQ-dW

DQ =0 by definition

Therefore, 0=dQ=dU +dW

The word done dW for the change in volume V by dV is given as PdV.

The first term is specific heat which is defined as the heat added per unit temperature change per mole of a substance. The heat that is added increases the internal energy U such that it justifies the definition of specific heat at constant volume is given as:

Cv =dUdT1n

Where,

n: number of moles

Therefore, 0= nCvdT+PdV …..(eq.1)

From the ideal gas law, we have

NRT =PV …..(eq.2)

Therefore, nRdT = PdV+VdP…. (eq.3)

By combining the equation 1. And equation 2, we get

−PdV= nCvdT = Cv / R (PdV+VdP)

0 = (1+Cv / R) PdV+Cv / RVdP

0= R+ Cv / Cv (dV / V) +dP /P

When the heat is added at constant pressure Cp, we have

Cp=Cv +R

0=γ (dV / V)+dP / P

Where the specific heat ɣ is given as:

γ ≡ Cp /Cv

From calculus, we have,

d (lnx)=dx / x

0=γd (lnV) + d (lnP)

0=d (γlnV+lnP) =d (lnPVγ)

PVγ=constant

Question 9) Explain the first law of thermodynamics?

Answer: sometimes our bodies start to sweat and feel warm when we are in a room full of people and the sweating becomes excessive if the room size is small. This happens because our body is trying to cool off hence heat transfers from our body in the form of ‘sweat’. This entails the first law of thermodynamics.

The first law of thermodynamics states that the total energy of an isolated system is constant. Energy can be transformed from one form to another, but can neither be created nor destroyed.

According to this law, some of the heat given to the system is used to change the internal energy while the rest is doing work by the system. Mathematically,

ΔQ=ΔU+ΔW

Where,

ΔQ = Heat supplied to the system

ΔW= Work done by the system.

ΔU = Change in the internal energy of the system.

If Q is positive, then there is a net heat transfer into the system, if W is positive, then there is work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. It can also be represented as ΔU=ΔQ−W

We can say that internal energy tends to increase when the heat is given to the system and vice versa.

Question 10) What are the advantage and disadvantages of the air conditioner?

Answer:

Advantages

Air conditioning can act as a form of protection in your home. It doesn’t just cool the indoor air down, improves the air quality, and prevents respiratory condition flair ups.

Let’s take a look at two of the most important advantages of air conditioning.

a) Improves the Air Quality

Your air conditioning system can filter harmful pollutants out of the air. That means things like bacteria, dust, pollen, and allergens are reduced inside the home. If you or your family has any kind of respiratory problem, clean air can minimize the risk of asthma attacks and allergies.

Air conditioning can also lower the humidity levels in the air. Lower humidity means less chance of mold and mildew growth. It also prevents odors and stale air from gathering in your house.

b) Prevents Heat Stroke

When you’re exposed to hot temperatures for too long, your body loses water. This can cause dehydration and lead to heat strokes.

Air conditioning helps prevent sweating, which lowers the amount of water your body loses when it gets hot. It also helps your body regulate its internal temperature and prevent heatstroke.

Heatstroke and dehydration can be serious conditions. In some cases, not getting medical help in time can result in brain and organ damage. Air conditioning keeps your home cool and reduces your risk of overheating.

Disadvantages

Air conditioners don’t have a lot of disadvantages. The few it does come with can be prevented with the right care. But if you don’t give your air conditioning unit the proper maintenance, you could end up with a few problems on your hands.

Here’s a quick look at the disadvantages of air conditioning.

a) Dry Skin

If you run your air conditioner too high for too long, your skin will lose some of its natural moisture. This can make your skin feel dry and irritated.

But there are a few ways to avoid this problem.

Don’t spend prolonged periods in high air conditioning. Turn the temperature down and use floor and ceiling fans to make the room cooler. If you like high air conditioning, you can always use lotion to keep your skin moisturized and comfortable.

b) Dirty Filters Can Be a Health Risk

Over time, the filter inside your air conditioner will get full of dust, dirt, hair follicles, bacteria, and other allergens. If you don’t clean your filters, their air will carry all that gunk through the rest of the house.

This can aggravate respiratory problems like allergies and asthma. As long as you clean your filters every few months, you shouldn’t experience any of these problems.

UNIT 2

GAS LAW AND GAS PROCESSES

Answer:

Given,

Initial pressure, P1 = 4kPa

Final pressure, P2 = 10kPa

Final volume, V2 = 12L

According to Boyle’s law, V1 = (P2V2)/P1

V1 = (10 kPa * 12 L)/4 kPa = 30 L

Therefore, the volume of container 1 is 30 L.

Question 2) a gas occupies a volume of 500cm3 at 0°C and 780 mm Hg. What volume (in liters) will it occupy at 80°C and 780 mm Hg?

Answer: Given, V1= 500 cm³

                                               V2 =?

                                                T1= 0°C= 0+273 = 273 K

                                                T2= 90°C= 90+273 = 363 K

Here the pressure is constant and only the temperature is changed.

Using Charles Law,

(V1/ T1)= (V2 / T2)

500 / 273=V2 / 363

V2=500 * 363 / 273

V2= 664.83 cm3

1 cubic centimetre = 0.001 litre =1 x 10-3 litre

∴ 664.83 cubic centimetre = 664.83 x 10-3 = 0.664 litres

Answer: Given,

The initial amount of air (n1) = 20 mol

The initial volume of the tyre (V1) = 60 L

The final volume of the tyre (V2) = 30 L

According to Avogadro’s law, the final amount of air in the tyre (n2) = (V2n1)/V1

= 30L * 20 moles / 60 L

The deflated tyre would contain 10 moles of air.

Question 4) Write the applications of the first law of thermodynamics to study the flow process?

Answer: Application of first law to study flow process

A steady ﬂow is one that remains unchanged with time, and therefore a steady ﬂow has the following characteristics:

Ii. Since the system remains unchanged with time during a steady ﬂow process, the system boundary also remains the same

Iv. Rates at which heat and work are transferred across the boundary of the system remain unchanged.

The application of the first law of thermodynamics to steady-flow processes is defined by two fundamental principles:

(a) Steady mass flow rate m = pAu (mass continuity) where p = fluid density

A = duct area

u = fluid velocity

And m = steady mass flow rate (dm/dt)

(This is usually recast as mv = uA since v = 1/p.)

(b) Conservation of energy (the steady-flow energy equation or SFEE)

Q 1-2 –W 1-2 =m [(h 2 -h 1) + ½ (u 2 2 –u 1 2) + g(Z 2 -Z l )]

Where Z = height above datum.

Note that very often gz is negligible compared with other terms and the reduced SFEE is written as

Q 2 -W 2 =m [(h2 –h1) + ½ (u 2 2 –u 1 2 ) ]

In per unit mass flow rate.

Q/m=q, W/m= w

q 2 - w 2 = (h 2 – h 1) + ½ (u 2 2 –u 1 2 )

An even further reduction by negligible kinetic energy terms here

q2 - w2 = (h2 – h1)

Many processes, in reality, approximate closely to a steady flow, for example, steady

Conditions in a steam power plant after the start-up transient is over and

Steady conditions exist at all points in the system.

Here two essential characteristics of steady flow:

(a) m is constant,

(b) The properties at any station are invariable with time.

The one obvious application of the SFEE where gz is prominent is in a hydroelectric plant where a change in z is essential for power production.

Question 5) Explain the difference between isothermal and adiabatic processes?

Answer: The Difference Between Isothermal and Adiabatic Process

Isothermal process	Adiabatic process
An isothermal process is defined as one of the thermodynamic processes which occur at a constant temperature	An adiabatic process is defined as one of the thermodynamic processes which occur without any heat transfer between the system and the surrounding
Work done is due to the change in the net heat content in the system	Work done is due to the change in its internal energy
The temperature cannot be varied	The temperature can be varied
There is a transfer of heat	There is no transfer of heat

Question 6) writes refrigerant properties?

Answer: Refrigerant properties:

Following are the refrigerant properties shown below

1) Low boiling point and Low freezing point.

2) Low specific heat and High latent heat.

3) High critical pressure and temperature

4) Low specific volume to reduce the size of the compressor.

5) High thermal conductivity to compact evaporator and condenser.

6) Non-flammable, non-explosive, non-toxic, and non-corrosive.

7) High miscibility with lubricating oil

8) High COP in the working temperature range.

9) Compatible with a legal requirement

10) Availability and cost

Question 7) Write the Difference between split and packaged units?

Answer: Difference between split and packaged unit

Packaged air conditioning units consume more power in comparison to the split unit.

The air handler of the split systems is installed indoors to enhance the life-span of the AC filters and evaporator coils.

Question 8) Explain the adiabatic process formula?

Answer: Adiabatic process derivation

The adiabatic process can be derived from the first law of thermodynamics relating to the change in internal energy dU to the work dW done by the system and the heat dQ added to it.

DU =dQ-dW

DQ =0 by definition

Therefore, 0=dQ=dU +dW

The word done dW for the change in volume V by dV is given as PdV.

Cv =dUdT1n

Where,

n: number of moles

Therefore, 0= nCvdT+PdV …..(eq.1)

From the ideal gas law, we have

NRT =PV …..(eq.2)

Therefore, nRdT = PdV+VdP…. (eq.3)

By combining the equation 1. And equation 2, we get

−PdV= nCvdT = Cv / R (PdV+VdP)

0 = (1+Cv / R) PdV+Cv / RVdP

0= R+ Cv / Cv (dV / V) +dP /P

When the heat is added at constant pressure Cp, we have

Cp=Cv +R

0=γ (dV / V)+dP / P

Where the specific heat ɣ is given as:

γ ≡ Cp /Cv

From calculus, we have,

d (lnx)=dx / x

0=γd (lnV) + d (lnP)

0=d (γlnV+lnP) =d (lnPVγ)

PVγ=constant

Question 9) Explain the first law of thermodynamics?

The first law of thermodynamics states that the total energy of an isolated system is constant. Energy can be transformed from one form to another, but can neither be created nor destroyed.

According to this law, some of the heat given to the system is used to change the internal energy while the rest is doing work by the system. Mathematically,

ΔQ=ΔU+ΔW

Where,

ΔQ = Heat supplied to the system

ΔW= Work done by the system.

ΔU = Change in the internal energy of the system.

We can say that internal energy tends to increase when the heat is given to the system and vice versa.

Question 10) What are the advantage and disadvantages of the air conditioner?

Answer:

Advantages

Air conditioning can act as a form of protection in your home. It doesn’t just cool the indoor air down, improves the air quality, and prevents respiratory condition flair ups.

Let’s take a look at two of the most important advantages of air conditioning.

a) Improves the Air Quality

Air conditioning can also lower the humidity levels in the air. Lower humidity means less chance of mold and mildew growth. It also prevents odors and stale air from gathering in your house.

b) Prevents Heat Stroke

When you’re exposed to hot temperatures for too long, your body loses water. This can cause dehydration and lead to heat strokes.

Air conditioning helps prevent sweating, which lowers the amount of water your body loses when it gets hot. It also helps your body regulate its internal temperature and prevent heatstroke.

Disadvantages

Here’s a quick look at the disadvantages of air conditioning.

a) Dry Skin

If you run your air conditioner too high for too long, your skin will lose some of its natural moisture. This can make your skin feel dry and irritated.

But there are a few ways to avoid this problem.

b) Dirty Filters Can Be a Health Risk

This can aggravate respiratory problems like allergies and asthma. As long as you clean your filters every few months, you shouldn’t experience any of these problems.

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