Thermodynamics MCQ Quiz - Objective Question with Answer for Thermodynamics - Download Free PDF

Calculation:

Given that 

\(n=1\ mole, W=6R\ J, \gamma=\frac{5}{3}\)

\( W =\frac{R\left ( T_{i}-T_{f} \right )}{\gamma -1} \)

\(\Rightarrow 6R = \frac{R\left ( T-T_{f} \right )}{\left ( \frac{5}{3}-1 \right )}\)

\(\Rightarrow T_{f} = \left ( T-4 \right ) K\)

Given: \(\gamma=7/5\)

According to Adiabatic process

Volume of the gas \( V = \dfrac{m}{d} \) and using \( PV^{\gamma} \) = constant \(\dfrac{P'}{P}= \left ( \dfrac{V}{V} \right )^{\gamma}= \left ( \dfrac{d}{d} \right )^{\gamma}\left ( 32 \right )^{7/5} = 128\)

isothermal processes are necessarily slow as they require heat transfer to remain at the same temperature which is done by being in thermal equilibrium with some reservoir. A process will be isothermal only if it happens on timescales larger than the timescale required for effective heat transfer.

The correct answer is 10.44.

Key Points

• The coefficient of performance (COP) of a refrigerator is given by the formula COP = T_C / (T_H - T_C).
• Here, T_C is the temperature of the cold reservoir (inside the refrigerator) and T_H is the temperature of the hot reservoir (room temperature).
• The temperatures must be converted to Kelvin before using the formula. T_C = 9ºC + 273 = 282K and T_H = 36ºC + 273 = 309K.
• Using the formula: COP = 282 / (309 - 282) = 282 / 27 = 10.44.

Important Points

• The coefficient of performance (COP) is a measure of the efficiency of a refrigerator or heat pump.
• The higher the COP, the more efficient the refrigerator is.
• In this context, efficiency means how well the refrigerator can transfer heat from the inside to the outside with less work input.
• It is important to always convert temperatures to Kelvin when using thermodynamic equations.

Concept and Explanation:

1 → 2 : Isobaric because Pressure is constant.

2 → 3 : adiabatic because entropy is constant throughout the process, indicating a reversible adiabatic process.

Δ S = ∫ dq /T

3 → 1 :  Isochoric because Volume is constant, as seen in P-V graph.

Some area will be enclosed in P-V graph cycle, hence work done is non-zero.

By first law of thermodynamics

Δ Q = Δ U + W

Since, ΔU is 0 for cyclic process, so

 Δ Q = W

Hence, heat absorbed is also non zero.

∴ Option 2),3) and 4) are correct

Concept:

• The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

This law is the basis for the temperature measurement.

• The first law of thermodynamics states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.

 

The first law of thermodynamics is a restatement of the law of conservation of energy

i.e., According to the first law of Thermodynamics:

ΔQ = ΔW + ΔU

• Now the First Law of Thermodynamics helped us in understanding the principle of conservation of energy, whereas according to the Second Law of thermodynamics for natural system heat always flows in one direction (higher temperature to lower temperature body) unless it aided by an external factor.

 

And to measure the direction of force we use term entropy which can be expressed as

\({\rm{\Delta }}S = \;\smallint \frac{{dQ}}{T}\)

ΔQ = heat exchange

ΔW = work done due to expansion

ΔU = internal energy of the system

ΔS = change in entropy

T = temperature

Explanation:

As explained above according to the first law of thermodynamics energy cannot be created or destroyed in an isolated system, energy can only be transferred or changed from one form to another.

This is the ideal statement which is used in thermodynamics for explaining the concept of energy conservation among systems and surrounding.

Hence option 2 is correct among all

Tricks to remember:

This is the conclusive point for all three laws of thermodynamics.

Zeroth law – Concept of temperature

First law – Concept of internal energy/ energy conservation

Second law – Concept of entropy/ heat flow

CONCEPT:

• Carnot engine: The theoretical engine which works on the Carnot cycle is called a Carnot engine.
  • It gives the maximum possible efficiency among all types of heat engines.
  • The part of the Carnot engine which provides heat to the engine is called a heat source.
  • The temperature of the source is maximum among all the parts.
  • The part of the Carnot engine in which an extra amount of heat is rejected by the engine is called as a heat sink.
  • The amount of work which is done by the engine is called as work done.

The efficiency (η) of a Carnot engine is given by:

\(η = 1 - \frac{{{T_C}}}{{{T_H}}} = \;\frac{{Work\;done\left( W \right)}}{{{Q_{in}}}} = \;\frac{{{Q_{in}} - \;{Q_R}}}{{{Q_{in}}}}\)

Where TC is the temperature of the sink, TH is the temperature of the source, W is work done by the engine, Qin is the heat given to the engine/heat input and QRis heat rejected.

EXPLANATION:

The efficiency (η) of the Carnot engine is given by:

η = 1 - T_C/T_H

• Here if T_H increases, the value of T_C/T_H decreases, and hence the value of (1 - TC/TH) increases.
• If the temperature of the source (T_H) is increased then the efficiency of the Carnot engine increases. So option 1 is correct.

CONCEPT:

• Carnot engine an ideal reversible engine that operates between two temperatures T1 (Source), and T2 (Sink).
• Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
• The steps of the Carnot cycle are
  1. Isothermal expansion
  2. Adiabatic expansion
  3. Isothermal compression
  4. Adiabatic compression
• The efficiency of the Carnot engine is defined as the ratio of net work done per cycle by the engine to heat absorbed per cycle by the working substance from the source. 
• The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

​\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

SOLUTION:

CONCEPT:

Isobaric process	Isochoric process	Adiabatic process	Isothermal process
It allows to set up the relationship between the changes in volume and temperature under constant pressure.	The process in which the volume of the gas remains constant is called the isochoric process.  For example: A gas is filled in a closed container then the volume of the gas will remain constant.	The thermodynamic process in a system, during which no heat transfer occurs between thermodynamic systems and surrounding is called an adiabatic process.	It allows to set up the relationship between the changes in pressure and volume under constant temperature:
V1/T1 = V2/T2 so V ∝ T   Where [V1 and V2 are volume and T1 and T2 are different temperatures]	\(\frac{P_1}{T_1} = \frac{P_2}{T_2} = Constant\)	PVγ  = Constant Where γ is ratio of specific heat	P1V1 = P2V2 so P V = Constant     Where [P1 and P2 are the pressure of gases and V1 and V2 are volume]

EXPLANATION:

• In an adiabatic process, no heat flows between the system and surroundings. So option 3 is correct.

CONCEPT:

• Carnot engine an ideal reversible engine that operates between two temperatures T₁ (Source), and T₂ (Sink).
• Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
• The steps of the Carnot cycle are
  1. Isothermal expansion
  2. Adiabatic expansion
  3. Isothermal compression
  4. Adiabatic compression
• The efficiency of the Carnot engine is defined as the ratio of network done per cycle by the engine to heat absorbed per cycle by the working substance from the source. 
• The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q₁ = Amount of heat absorbed, Q₂ = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

​\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

CALCULATION :

Given - T1 = 227+273 = 500 K, T₂ = 127 +273 = 400 K, Heat absorbed by the engine is Q₁ = 6 × 10⁴J

• The efficiency of the heat engine is given by

\(\Rightarrow \eta = \frac{W}{Q_1} = 1-\frac{T_2}{T_1}\)

\(\Rightarrow \frac{W}{6\times 10^{4}} = 1-\frac{400}{500}\)

\(\Rightarrow \frac{W}{6\times 10^{4}} = 1-\frac{4}{5} \)

\(\Rightarrow \frac{W}{6\times10^{4}} = \frac{1}{5} \)

\(\Rightarrow {W} = \frac{6\times 10^{4}}{5} = 1.2 \times 10^{4} J\)

• Hence, option 4 is the answer

CONCEPT:

• State variables are defined as the thermodynamical variables which depend only on the initial and final state of a thermodynamical system. 
  • These variables don't depend on how the thermodynamical system changed itself from the initial to the final state.
  • Temperature, Pressure, Internal energy, and Density are the examples of state variables.
  • State variables are also known as state functions.
• Path variables are defined as the thermodynamical variables which depend on the way in which the thermodynamical system achieved the initial and final states.
  • Heat, Work is examples of Path variables 

 

EXPLANATION:

• Internal energy, pressure, density, and enthalpy are examples of state variables. Since they depend only on the final and initial states of the thermodynamical system.
• Heat is a measure of the amount of energy present in a thermodynamical system. As the amount of energy changes the heat present in the system changes. Hence heat is the path variable. Therefore option 4 is correct answer.

CONCEPT:

Charles law:

• If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

i.e. V ∝ T
or V/T = constant
\( \Rightarrow \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\)

EXPLANATION:

• From above it is clear that in Charles's Law, pressure remains constant. Therefore option 4 is correct.

CONCEPT:

• Carnot engine: A theoretical thermodynamic cycle proposed by Leonard Carnot. It gives the estimate of the maximum possible efficiency that a heat engine during the conversion process of heat into work and conversely, working between two reservoirs, can possess.
  • So practically and theoretically there can not be any engine with more efficiency than Carnot engine.

The efficiency of the Carnot's Heat engine is given by:

 \(η=1-\frac{T_c}{T_h}\)

where T_c is the temperature of the cold reservoir/sink and T_h is the temperature of the hot reservoir/source.

• The efficiency of this type of engine is independent of the nature of the working substance and is only dependent on the temperature of the hot and cold reservoirs.

CALCULATION:

Given that sink temperature T_c = 27°C = 300K

η = 70% = 0.7

\(η=1-\frac{T_c}{T_h}\)

\(0.7=1-\frac{300}{T_h}\)

T_h = 300/0.3 = 1000 K 

So the correct answer is option 1.

CONCEPT:

First Law of Thermodynamics:

• It is a statement of conservation of energy in the thermodynamical process.
• According to it heat given to a system (ΔQ) is equal to the sum of the increase in its internal energy (ΔU) and the work done (ΔW) by the system against the surroundings.

i.e ΔQ = ΔU + ΔW          [∴ ΔW = p ΔV]

• It makes no distinction between work and heat as according to it the internal energy (and hence temperature) of a system may be increased either by adding heat to it or doing work on it or both.

EXPLANATION:

• When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as an isothermal process.
• As we know that, the internal energy of the system is a function of temperature alone, so in the isothermal process, the change in internal energy is zero.

⇒ ΔQ = 0 + ΔW =  ΔW

• Therefore, the heat given to an ideal gas in isothermal conditions is used to do external work. Hence option 2 is correct.

Concept:

• The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.
• When heat energy is supplied to a thermodynamic system or any machine. 
• Two things may occur:
  • The internal energy of the System or machine may change.
  • The system may do some external work.

According to the first law of Thermodynamics:

ΔQ = ΔW + ΔU

Where ΔQ = Heat supplied to the system or heat exchange, ΔW = work done by the system, ΔU = change in internal energy of the system 

Explanation:

Given that, ΔQ = 110 J, ΔU = 40 J

According to the  first law of thermodynamics:

ΔQ = ΔW + ΔU

ΔU = ΔQ - ΔW

40 J = 110 J - ΔW

ΔW = 110 - 40 = 70 J

Then the amount of external work done is 70 J.