Class 12 Physics Chapter 4: First Law of Thermodynamics Complete Guide (NEB New Syllabus) | Notes, Formula, Numerical Problems & PDF
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1. Syllabus

Unit: Heat and Thermodynamics (16 hours)

Chapter 4: First Law of Thermodynamics (8 hours)

4.1 Thermodynamic systems

4.2 Work done during volume change

4.3 Heat and work; internal energy and first law of thermodynamics

4.4 Thermodynamic processes: adiabatic, isochoric, isothermal and isobaric

4.5 Heat capacities of an ideal gas at constant pressure and volume and relation between them

4.6 Isothermal and adiabatic processes for an ideal gas

These notes follow the NEB Class 12 Physics syllabus (+2 Physics, Nepal). Practise the multiple choice questions first, then work through the numericals, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. Which of the following is NOT a thermodynamic system?

  • a) Open system
  • b) Closed system
  • c) Isolated system
  • d) Static system

Correct Answer: d) Static system

Reason: Thermodynamic systems are classified by their ability to exchange mass and energy with the surroundings: open (both), closed (energy only) and isolated (neither). “Static system” is not a thermodynamic classification.


Question 2. The first law of thermodynamics is a statement of the conservation of:

  • a) Mass
  • b) Momentum
  • c) Energy
  • d) Entropy

Correct Answer: c) Energy

Reason: The first law, \(\Delta Q=\Delta U+\Delta W\), says the heat supplied to a system equals the change in its internal energy plus the work done by it. Energy is neither created nor destroyed, only transferred or transformed.


Question 3. If a gas expands adiabatically, which of the following statements is true?

  • a) The internal energy of the gas remains constant
  • b) The gas absorbs heat
  • c) The work done by the gas is equal to the decrease in internal energy
  • d) The temperature of the gas remains unchanged

Correct Answer: c) The work done by the gas is equal to the decrease in internal energy

Reason: In an adiabatic process \(\Delta Q=0\), so

\[ 0=\Delta U+\Delta W\;\Rightarrow\;\Delta W=-\Delta U \]

During expansion the gas does positive work (\(\Delta W>0\)), so its internal energy decreases (\(\Delta U<0\)) and it cools.


Question 4. In an isothermal process, which of the following remains constant?

  • a) Pressure
  • b) Temperature
  • c) Volume
  • d) Internal energy

Correct Answer: b) Temperature

Reason: “Iso-thermal” means constant temperature: the system exchanges heat slowly with a reservoir so that \(T\) stays fixed. For an ideal gas the internal energy depends only on temperature and therefore also stays constant, but constant temperature is the defining feature of the process.


Question 5. Which of the following correctly describes the relation between heat capacity at constant pressure and constant volume?

  • a) \(C_p=C_v\)
  • b) \(C_p<C_v\)
  • c) \(C_p-C_v=R\)
  • d) \(C_p-C_v=\gamma\)

Correct Answer: c) \(C_p-C_v=R\)

Reason: This is Meyer’s relation for an ideal gas. Heating at constant pressure lets the gas expand and do work against the surroundings, so \(C_p\) exceeds \(C_v\) by the universal gas constant \(R\).


Question 6. A gas undergoes an isochoric (constant volume) process. What is the work done during this process?

  • a) Positive
  • b) Negative
  • c) Zero
  • d) Cannot be determined

Correct Answer: c) Zero

Reason: \(W=\int P\,dV\). At constant volume \(dV=0\), so the work done is zero.

Question 7. A gas expands isothermally at a temperature of 300 K from an initial volume of 2 L to a final volume of 6 L. If the gas follows \(W=nRT\ln\!\left(\dfrac{V_f}{V_i}\right)\), the work done is:

  • a) Positive
  • b) Negative
  • c) Zero
  • d) Infinite

Correct Answer: a) Positive

Reason: The gas expands, so \(\dfrac{V_f}{V_i}=\dfrac62=3>1\) and \(\ln 3>0\). Hence \(W>0\): the gas does work on its surroundings.


Question 8. A gas undergoes an adiabatic compression. What happens to its temperature?

  • a) Increases
  • b) Decreases
  • c) Remains constant
  • d) Becomes zero

Correct Answer: a) Increases

Reason: With \(\Delta Q=0\), \(\Delta U=-\Delta W\). In compression work is done on the gas, so \(\Delta W<0\) and \(\Delta U>0\). The internal energy, and therefore the temperature, rises.


Question 9. 0.5 mol of an ideal gas expands isothermally and reversibly at \(27^\circ\mathrm{C}\) from 1 L to 4 L. The work done (in J) is closest to:

  • a) 1384 J
  • b) 1725 J
  • c) 1845 J
  • d) 2100 J

Correct Answer: b) 1725 J

Reason: \(T=27+273=300\ \mathrm{K}\).

\[ W=nRT\ln\frac{V_2}{V_1}=0.5\times8.314\times300\times\ln4=1247.1\times1.386\approx1729\ \mathrm{J} \]

The closest option is 1725 J.


Question 10. A sample of 2 moles of an ideal gas (\(\gamma=1.4\)) is compressed adiabatically from a volume of 10 L to 1 L. If the initial temperature is 300 K, what is the final temperature?

  • a) 750 K
  • b) 820 K
  • c) 900 K
  • d) 1050 K

Correct Answer: a) 750 K

Reason: For an adiabatic process \(T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\).

\[ T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=300\times(10)^{0.4}=300\times2.512\approx753\ \mathrm{K}\approx750\ \mathrm{K} \]

Question 11. A gas expands according to the relation \(PV^{1.4}=\text{constant}\). If the initial pressure and volume are 2 atm and 3 L respectively, find the final pressure when the volume becomes 6 L.

  • a) 0.57 atm
  • b) 0.63 atm
  • c) 0.76 atm
  • d) 1.2 atm

Correct Answer: c) 0.76 atm

Reason: Using \(P_1V_1^{1.4}=P_2V_2^{1.4}\):

\[ P_2=2\times\left(\frac36\right)^{1.4}=2\times(0.5)^{1.4}=2\times0.3789\approx0.76\ \mathrm{atm} \]

3. Numerical and Subjective Questions

Question 1. Thermodynamic systems and work done during volume change.

a) Define a thermodynamic system. Write one example of each type of thermodynamic system. Differentiate between open, closed and isolated systems with examples.

A thermodynamic system is a specific, macroscopic region of space or quantity of matter chosen for thermodynamic study, separated from its surroundings by a real or imaginary boundary.

Open system: exchanges both mass and energy (heat or work) with its surroundings. Example: boiling water in an open pan, where steam escapes and heat is transferred.

Closed system: exchanges energy but not mass. Example: gas sealed in a cylinder with a movable piston; energy is transferred as heat or by piston movement but the gas does not escape.

Isolated system: exchanges neither mass nor energy. Example: hot coffee in a perfectly insulated, sealed thermos flask.

b) How does work done on the system differ from work done by the system?

Work done by the system (expansion): the system pushes on the surroundings, for example a gas pushing a piston outward. \(dV\) is positive, so the work is taken as positive (\(+W\)) and the system loses internal energy to perform it.

Work done on the system (compression): the surroundings push on the system, for example a piston pushed inward to compress a gas. \(dV\) is negative, so the work is taken as negative (\(-W\)) and the internal energy of the system increases.

c) A gas at pressure \(2\times10^5\ \mathrm{Pa}\) and volume \(0.01\ \mathrm{m^3}\) expands isobarically to \(0.03\ \mathrm{m^3}\). Find the work done.

Given: \(P=2\times10^5\ \mathrm{Pa},\ V_1=0.01\ \mathrm{m^3},\ V_2=0.03\ \mathrm{m^3}\)

For a constant-pressure process:

\[ W=P(V_2-V_1)=(2\times10^5)(0.03-0.01)=4000\ \mathrm{J} \]

d) Define an indicator diagram and explain how work done by an expanding gas can be calculated from it.

An indicator diagram (P-V diagram) is a graph of the pressure \(P\) of a thermodynamic system on the y-axis against its volume \(V\) on the x-axis.

The work done by an expanding gas is

\[ W=\int_{V_1}^{V_2}P\,dV \]

which is equal to the area under the P-V curve between the initial volume \(V_1\) and the final volume \(V_2\). When the volume increases (left to right) the work is positive.


Question 2. Internal energy of a gas and its relation with heat and work.

a) Explain the concept of internal energy and its relation with heat and work. What does the equation \(dU=-dW\) signify? How does the internal energy of a real gas differ from that of an ideal gas?

Internal energy: internal energy \(U\) is the sum of the microscopic kinetic energies (translational, rotational, vibrational) and potential energies (due to intermolecular forces) of all the molecules of a system. It is a state function.

Relation with heat and work: the first law, \(dQ=dU+dW\), connects them. Heat added to a system raises its internal energy and does external work.

Meaning of \(dU=-dW\): it follows when \(dQ=0\), that is, in an adiabatic process. If the gas expands and does work (\(dW>0\)), the work is done at the expense of the internal energy, so \(dU\) is negative and the gas cools.

Ideal and real gas: an ideal gas has no intermolecular forces, so its internal energy is purely kinetic and depends only on temperature. A real gas has intermolecular forces, so its internal energy has both kinetic and potential parts and depends on temperature and volume.

b) Calculate the change in internal energy of 2 moles of a monoatomic ideal gas which undergoes a temperature change from 300 K to 500 K. (\(C_v=3R/2\))

Given: \(n=2\ \mathrm{mol},\ T_1=300\ \mathrm{K},\ T_2=500\ \mathrm{K},\ C_v=\tfrac{3R}{2},\ R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}\)
\[ \Delta U=nC_v(T_2-T_1)=2\times\frac{3\times8.314}{2}\times(500-300) \]
\[ \Delta U=3\times8.314\times200=4988.4\ \mathrm{J} \]

Question 3. The first law of thermodynamics.

a) State and explain the first law of thermodynamics. Discuss its significance.

Statement: if an amount of heat \(dQ\) is supplied to a system, it is used in two ways: to increase the internal energy of the system (\(dU\)) and to do external work (\(dW\)) by the system.

\[ dQ=dU+dW \]

Significance: it establishes the equivalence of heat and mechanical work, shows that heat is a form of energy, and states that every thermodynamic process obeys conservation of energy.

b) Does it follow the law of conservation of energy?

Yes. The first law is the principle of conservation of energy applied to thermodynamic systems. Energy can neither be created nor destroyed; it only changes from heat into internal energy and mechanical work.

c) A gas absorbs 600 J of heat and does 150 J of work. Find the change in internal energy.

Given: \(dQ=+600\ \mathrm{J}\) (heat absorbed), \(dW=+150\ \mathrm{J}\) (work done by the gas)
\[ dU=dQ-dW=600-150=450\ \mathrm{J} \]

d) List out the limitations of the first law of thermodynamics.

Direction of heat flow: it does not tell in which direction heat flows, for example why heat flows from hot to cold and not the other way.

Spontaneity: it gives no information on whether a given process is possible or spontaneous.

Extent of conversion: it does not say how much heat can be converted into work; the second law is needed to show that complete conversion in a cycle is impossible.

Question 4. Heat capacities of an ideal gas and Meyer’s relation.

a) Why do gases have two specific heat capacities?

Solids and liquids change very little in volume on heating, so one specific heat is enough. A gas, however, can change greatly in volume or pressure. Heated at constant volume, all the heat goes into internal energy. Heated at constant pressure, the gas also expands and does external work. Different amounts of heat are needed for the same temperature rise in the two cases, so a gas has two principal specific heats: \(C_v\) (constant volume) and \(C_p\) (constant pressure).

b) Derive Meyer’s relation, \(C_p-C_v=R\).

Consider 1 mole of an ideal gas.

At constant volume: \(dQ_v=C_v\,dT\). Since \(dW=P\,dV=0\), the first law gives \(dU=C_v\,dT\).

At constant pressure: \(dQ_p=C_p\,dT\), and the gas does work \(dW=P\,dV\). The first law gives

\[ C_p\,dT=dU+P\,dV \]

The internal energy of an ideal gas depends only on temperature, so \(dU=C_v\,dT\) is the same in both cases:

\[ C_p\,dT=C_v\,dT+P\,dV \]

From \(PV=RT\) at constant pressure, \(P\,dV=R\,dT\). Therefore

\[ C_p\,dT=C_v\,dT+R\,dT\]
\[ C_p-C_v=R \]

c) Why is \(C_p\) always greater than \(C_v\)?

At constant volume all the heat supplied raises the internal energy (temperature). At constant pressure the gas also expands and does external work, so extra heat must be supplied for this work in addition to the heat that raises the internal energy. Hence \(C_p>C_v\).

d) If \(C_p=29.1\ \mathrm{J\,mol^{-1}K^{-1}}\) and \(C_v=20.8\ \mathrm{J\,mol^{-1}K^{-1}}\), find the gas constant \(R\) and \(\gamma\).

\[ R=C_p-C_v=29.1-20.8=8.3\ \mathrm{J\,mol^{-1}K^{-1}} \]
\[ \gamma=\frac{C_p}{C_v}=\frac{29.1}{20.8}\approx1.4 \]

Question 5. Thermodynamic processes and their P-V diagrams.

a) Explain the processes: isothermal, adiabatic, isobaric and isochoric with the help of P-V diagrams.

Isothermal: temperature remains constant (\(\Delta T=0\)). It occurs slowly in a highly conducting vessel, and \(PV=\text{constant}\).

Adiabatic: no heat is exchanged with the surroundings (\(\Delta Q=0\)). It occurs rapidly or in a well-insulated vessel, and \(PV^{\gamma}=\text{constant}\).

Isobaric: pressure remains constant (\(\Delta P=0\)), for example heating a gas in a cylinder with a freely moving, weight-loaded piston.

Isochoric: volume remains constant (\(\Delta V=0\)), for example heating a gas in a rigid sealed container.

b) Differentiate between the different types of thermodynamic processes. Compare their P-V diagrams.

Starting from the same initial state \((P_1,V_1)\):

Class 12 Physics Chapter 4 First Law of Thermodynamics=

Figure: P-V diagrams of the four thermodynamic processes from the same initial state

Isobaric is a horizontal straight line (slope \(=0\)). Isochoric is a vertical straight line (slope \(=\infty\)). Isothermal is a rectangular hyperbola sloping downward with slope \(\dfrac{dP}{dV}=-\dfrac PV\). Adiabatic is also a downward curve but with slope \(\dfrac{dP}{dV}=-\gamma\dfrac PV\); since \(\gamma>1\), it is steeper than the isothermal curve.

ProcessConditionWork done \(W\)\(\Delta U\)Heat \(Q\)
Isothermal\(T\) constant\(nRT\ln\dfrac{V_2}{V_1}\)0\(Q=W\)
Adiabatic\(Q=0\)\(\dfrac{nR(T_1-T_2)}{\gamma-1}\)\(-W\)0
Isobaric\(P\) constant\(P(V_2-V_1)\)\(nC_v\Delta T\)\(nC_p\Delta T\)
Isochoric\(V\) constant0\(nC_v\Delta T\)\(Q=\Delta U\)

Question 6. Adiabatic process.

a) Define an adiabatic process. Derive the equation \(PV^\gamma=\text{constant}\).

An adiabatic process is a thermodynamic change in which no heat enters or leaves the system (\(dQ=0\)).

For \(n\) moles, the first law with \(dQ=0\) gives

\[ nC_v\,dT+P\,dV=0\qquad\cdots(1) \]

Differentiating \(PV=nRT\):

\[ P\,dV+V\,dP=nR\,dT\;\Rightarrow\;dT=\frac{P\,dV+V\,dP}{nR} \]

Substituting in (1) and using \(R=C_p-C_v\):

\[ \frac{C_v}{C_p-C_v}\,(P\,dV+V\,dP)+P\,dV=0 \]

Since \(\dfrac{C_v}{C_p-C_v}=\dfrac1{\gamma-1}\), multiplying through by \((\gamma-1)\) gives

\[ P\,dV+V\,dP+(\gamma-1)P\,dV=0\;\Rightarrow\;V\,dP+\gamma P\,dV=0 \]

Dividing by \(PV\) and integrating:

\[ \frac{dP}{P}+\gamma\frac{dV}{V}=0\;\Rightarrow\;\ln P+\gamma\ln V=\text{constant} \]
\[ PV^{\gamma}=\text{constant} \]

b) In an adiabatic expansion, a gas does 1200 J of work. Find the change in internal energy.

Adiabatic, so \(Q=0\), and the work done by the gas is \(W=+1200\ \mathrm{J}\):

\[ 0=\Delta U+1200\;\Rightarrow\;\Delta U=-1200\ \mathrm{J} \]

c) Find the final temperature and the work done during an adiabatic expansion of 1 mole of ideal gas from 1 L to 5 L. Take \(\gamma=1.4\), \(T_1=300\ \mathrm{K}\), \(R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}\).

Final temperature:

\[ T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\;\Rightarrow\;T_2=300\times\left(\frac15\right)^{0.4}=\frac{300}{1.9037}\approx157.6\ \mathrm{K} \]

Work done:

\[ W=\frac{nR(T_1-T_2)}{\gamma-1}=\frac{1\times8.314\times(300-157.6)}{0.4}\approx2960\ \mathrm{J} \]

Question 7. Work done in isothermal and adiabatic processes.

a) Derive the expression for the work done during an isothermal and an adiabatic process.

Isothermal work. \(W=\displaystyle\int_{V_1}^{V_2}P\,dV\) with \(P=\dfrac{nRT}{V}\) and constant \(T\):

\[ W=nRT\int_{V_1}^{V_2}\frac{dV}{V}=nRT\ln\frac{V_2}{V_1}=2.303\,nRT\log_{10}\frac{V_2}{V_1} \]

Adiabatic work. For an adiabatic process \(PV^\gamma=K\), so \(P=KV^{-\gamma}\):

\[ W=K\int_{V_1}^{V_2}V^{-\gamma}dV=\frac{K\left(V_2^{1-\gamma}-V_1^{1-\gamma}\right)}{1-\gamma} \]

Using \(K=P_1V_1^\gamma=P_2V_2^\gamma\):

\[ W=\frac{P_2V_2-P_1V_1}{1-\gamma}=\frac{P_1V_1-P_2V_2}{\gamma-1}=\frac{nR(T_1-T_2)}{\gamma-1} \]

b) A gas expands isothermally from 2 L to 8 L at 300 K. Calculate the work done. (\(R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}\), \(n=1\ \mathrm{mol}\))

\[ W=nRT\ln\frac{V_2}{V_1}=1\times8.314\times300\times\ln4=2494.2\times1.3863\approx3457.7\ \mathrm{J} \]

Question 8. Heat capacity ratio and reversibility.

a) Define the heat capacity ratio (\(\gamma\)).

The heat capacity ratio is the ratio of the molar heat capacity of a gas at constant pressure to that at constant volume:

\[ \gamma=\frac{C_p}{C_v}>1 \]

b) How does \(\gamma\) vary with the atomicity of a gas? Explain using degrees of freedom.

For a gas with \(f\) degrees of freedom, \(\gamma=1+\dfrac2f\).

Type of gasMotion\(f\)\(\gamma\)
Monoatomic (He, Ar)Translational only3\(\dfrac53\approx1.67\)
Diatomic (O₂, N₂)Translational and rotational5\(\dfrac75=1.4\)
Polyatomic (CO₂, NH₃)Complex rotation6 (typically)\(\dfrac43\approx1.33\)

As atomicity, and with it the degrees of freedom, increases, \(\gamma\) decreases.

c) Differentiate between a reversible and an irreversible process.

Reversible process: an idealised process that occurs infinitely slowly so that the system is in thermal and mechanical equilibrium with its surroundings at every stage. It can be retraced along the same path, leaving no net change in the system or the universe.

Irreversible process: a real process occurring at a finite rate, usually involving friction or rapid heat transfer. The system is not in equilibrium at each step and cannot be restored to its original state without a permanent change in the surroundings (an increase in entropy). All natural processes are irreversible.

4. View PDF Notes

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5. Frequently Asked Questions

What is the first law of thermodynamics? (definition)
The first law of thermodynamics states that when heat dQ is supplied to a system, it is used partly to increase the internal energy of the system (dU) and partly to do external work (dW). It is the law of conservation of energy applied to thermodynamic systems: energy can be transformed from one form to another but can neither be created nor destroyed.
What is the formula of the first law of thermodynamics?
In the physics (NEB) convention the formula is ΔQ = ΔU + ΔW, where ΔQ is the heat supplied to the system, ΔU is the change in internal energy and ΔW is the work done by the system. In differential form dQ = dU + dW, with dW = P dV. In many chemistry books the same law is written ΔU = q + w, where w is the work done on the system, so only the sign convention for work differs.
What is the first law of thermodynamics in Physics?
In physics the first law connects heat, internal energy and mechanical work: Q = ΔU + W. It explains why a gas cools when it expands adiabatically (Q = 0 so W = -ΔU), why the heat absorbed in an isothermal expansion equals the work done (ΔU = 0 for an ideal gas), and why Cp is greater than Cv. It does not tell the direction of a process, which needs the second law.
What is the first law of thermodynamics in chemistry?
In chemistry the first law is usually written ΔU = q + w, where q is the heat absorbed by the system and w is the work done on the system. It is used to find energy changes in reactions: at constant volume the heat exchanged equals ΔU, and at constant pressure the heat exchanged equals the enthalpy change ΔH = ΔU + PΔV.
What is the first law of thermodynamics in biology?
In biology the first law means that living organisms cannot create energy; they only convert it from one form to another. Plants convert light energy into chemical energy in photosynthesis, and cells convert the chemical energy stored in food into ATP, mechanical work and heat during respiration. The total energy stays constant.
What is the second law of thermodynamics?
The second law of thermodynamics gives the direction of natural processes. The Clausius statement says heat cannot flow by itself from a colder body to a hotter body, and the Kelvin-Planck statement says no heat engine can convert all the heat it absorbs completely into work in a cycle. Equivalently, the entropy of an isolated system never decreases. The first law cannot explain these limits, which is why the second law is needed.
What topics are included in Chapter 4 First Law of Thermodynamics of NEB Class 12 Physics?
The chapter (8 hours, part of the Heat and Thermodynamics unit of 16 hours) covers: 4.1 thermodynamic systems; 4.2 work done during volume change; 4.3 heat and work, internal energy and the first law; 4.4 adiabatic, isochoric, isothermal and isobaric processes; 4.5 heat capacities of an ideal gas at constant pressure and volume and the relation between them; 4.6 isothermal and adiabatic processes for an ideal gas.
What are open, closed and isolated thermodynamic systems?
An open system exchanges both mass and energy with its surroundings (boiling water in an open pan). A closed system exchanges energy but not mass (gas sealed in a cylinder with a movable piston). An isolated system exchanges neither mass nor energy (hot coffee in a perfectly insulated sealed flask).
How is work done during volume change calculated?
The work done by a gas is W = ∫P dV from the initial volume V₁ to the final volume V₂. On a P-V indicator diagram it equals the area under the curve. At constant pressure W = P(V₂ – V₁); for example, 2 × 10⁵ Pa expanding from 0.01 m³ to 0.03 m³ does 4000 J of work. At constant volume the work done is zero.
What does the first law say for isothermal, adiabatic, isobaric and isochoric processes?
Isothermal (ΔU = 0 for an ideal gas): Q = W. Adiabatic (Q = 0): W = -ΔU, so an expanding gas cools and a compressed gas heats up. Isobaric (P constant): Q = ΔU + PΔV. Isochoric (V constant, W = 0): Q = ΔU = nCvΔT.
What is Meyer’s relation and why is Cp greater than Cv?
Meyer’s relation for an ideal gas is Cp – Cv = R. Cp is greater than Cv because at constant pressure the gas expands and does external work, so extra heat is needed on top of the heat that raises the internal energy. The ratio γ = Cp/Cv is greater than 1: 5/3 for monoatomic, 7/5 for diatomic and about 4/3 for polyatomic gases.
What are the equations for isothermal and adiabatic processes of an ideal gas?
Isothermal: PV = constant and W = nRT ln(V₂/V₁). Adiabatic: PV^γ = constant, TV^(γ-1) = constant and W = nR(T₁ – T₂)/(γ – 1) = (P₁V₁ – P₂V₂)/(γ – 1). The adiabatic curve on a P-V diagram is steeper than the isothermal curve because γ is greater than 1.
What are the limitations of the first law of thermodynamics?
The first law does not give the direction of heat flow, does not say whether a process is spontaneous or even possible, and does not state how much heat can be converted into work. These limits are explained by the second law of thermodynamics.
Where can I get Class 12 Physics First Law of Thermodynamics notes in PDF?
This page includes handwritten notes and typed PDF notes for the First Law of Thermodynamics that you can view online, along with solved multiple choice questions and numericals.

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