Question 1

   Question: Two gases $A$ and $B$ are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas $A$ and $B$ are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio $\frac{r_A}{r_B}$ is equal to.

   Options: 

    A. $\frac{2}{\sqrt{3}}$

    B. $\frac{\sqrt{3}}{2}$

    C. $\frac{4}{3}$

    D. $\frac{3}{4}$

   Correct Answer: D

   Year: NEET 2025

   Solution: The problem involves two gases at the same pressure where equal amounts of heat are supplied and the internal energy change is the same. Applying the first law of thermodynamics, $\Delta Q = \Delta U + P \Delta V$, we find that since $\Delta Q$ and $\Delta U$ are equal for both, their work done ($W = P \Delta V$) must also be equal,. Using the piston area and displacement, we get $\pi r_A^2 d_A = \pi r_B^2 d_B$, which leads to the ratio $\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$,.

   Step Solution:

    1.  Apply First Law: State $\Delta Q = \Delta U + W$. Given $\Delta Q_A = \Delta Q_B$ and $\Delta U_A = \Delta U_B$, then $W_A = W_B$,.

    2.  Define Work: At constant pressure $P$, $W = P \Delta V$, so $P \Delta V_A = P \Delta V_B$, which simplifies to $\Delta V_A = \Delta V_B$.

    3.  Relate Volume to Geometry: Volume change $\Delta V$ is the area of the piston ($\pi r^2$) multiplied by displacement ($d$), giving $\pi r_A^2 d_A = \pi r_B^2 d_B$.

    4.  Isolate Ratio: Rearrange the equation to find $\frac{r_A^2}{r_B^2} = \frac{d_B}{d_A}$.

    5.  Final Calculation: $\frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}$,.

   Difficulty Level: Medium.

   Concept Name: First Law of Thermodynamics / Work Done at Constant Pressure,.

   Short cut solution: Since $\Delta Q$ and $\Delta U$ are constant, $W$ is constant. For constant $P$, $\Delta V$ must be constant. Because $\Delta V \propto r^2 d$, the ratio is simply $r \propto \frac{1}{\sqrt{d}}$.

Question 27

   Question: A sample of 0.1g of water at $100^\circ C$ and normal pressure $(1.013 \times 10^5 Nm^{-2})$ requires 54 cal of heat energy to convert to steam at $100^\circ C$. If the volume of the steam produced is 167.1cc, the change in internal energy of the sample, is.

   Options: 

    A. 104.3J

    B. 208.7J

    C. 42.2J

    D. 84.5J

   Correct Answer: B

   Year: NEET 2018

   Solution: $\Delta Q = \Delta U + P \Delta V$. $54 \times 4.18 = \Delta U + 1.013 \times 10^5 (167.1 \times 10^{-6} - 0)$. Solving this gives $\Delta U = 208.7 J$.

   Step Solution:

    1.  Convert Heat to Joules: $Q = 54 \text{ cal} \times 4.18 \text{ J/cal} \approx 225.7 \text{ J}$.

    2.  Calculate Work Done: $W = P \Delta V = (1.013 \times 10^5 \text{ Pa}) \times (167.1 \times 10^{-6} \text{ m}^3) \approx 16.9 \text{ J}$.

    3.  Apply First Law: $\Delta U = Q - W$.

    4.  Substitute Values: $\Delta U = 225.7 \text{ J} - 16.9 \text{ J}$.

    5.  Final Result: $\Delta U = 208.7 \text{ J}$.

   Difficulty Level: Medium.

   Concept Name: First Law of Thermodynamics / Latent Heat and Work.

   Short cut solution: Directly plug values into the energy balance equation: $\Delta U = (54 \times 4.18) - (1.013 \times 0.1671) \times 10^2$.

Question 44

   Question: Figure shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be.

   Options: 

    A. 460 J

    B. 300 J

    C. 380 J

    D. 500 J

   Correct Answer: A

   Year: 2015

   Solution: Since initial and final states are same, $\Delta U_{ABC} = \Delta U_{AC}$. In path AB (isochoric), $\Delta W_{AB}=0$ and $\Delta U_{AB}=400J$. In path BC, $W_{BC} = 120J$ and $\Delta U_{BC} = -20J$. Total $\Delta U_{AC} = 380J$. For path AC, $W_{AC} = 80J$, so $Q_{AC} = 380 + 80 = 460J$.

   Step Solution:

    1.  Internal Energy for Path AB: Since volume is constant, $W_{AB}=0$, so $\Delta U_{AB} = Q_{AB} = 400 \text{ J}$.

    2.  Internal Energy for Path BC: Calculate $W_{BC} = P \Delta V = 6 \times 10^4 \times 2 \times 10^{-3} = 120 \text{ J}$. Then $\Delta U_{BC} = Q_{BC} - W_{BC} = 100 - 120 = -20 \text{ J}$.

    3.  Total State Change: $\Delta U_{AC} = \Delta U_{AB} + \Delta U_{BC} = 400 - 20 = 380 \text{ J}$.

    4.  Work for Path AC: Area under the graph for path AC is $(2 \times 10^4 \times 2 \times 10^{-3}) + \frac{1}{2}(2 \times 10^{-3} \times 4 \times 10^4) = 40 + 40 = 80 \text{ J}$.

    5.  Calculate Heat for Path AC: $Q_{AC} = \Delta U_{AC} + W_{AC} = 380 + 80 = 460 \text{ J}$.

   Difficulty Level: Hard.

   Concept Name: First Law of Thermodynamics / Path Independence of Internal Energy.

   Short cut solution: Use the cyclic property that total heat in a cycle equals total work, or in this case, equate the internal energy changes of different paths between the same two points.

Question 52

   Question: A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is $U_a = 10\text{ J}$. Along the path adc the amount of heat absorbed $Q_1 = 50\text{ J}$ and the work obtained $W_1 = 20\text{ J}$ whereas along the path abc the heat absorbed $Q_2 = 36\text{ J}$. The amount of work along the path abc is.

   Options: 

    A. $10\text{ J}$

    B. $12\text{ J}$

    C. $36\text{ J}$

    D. $6\text{ J}$

   Correct Answer: D.

   Year: KN NEET 2013.

   Solution: According to the first law of thermodynamics, $\delta Q = \delta U + \delta W$. Along the path adc, the change in internal energy is $\delta U_1 = \delta Q_1 - \delta W_1 = 50\text{ J} - 20\text{ J} = 30\text{ J}$. Since the change in internal energy is path independent, $\delta U_1 = \delta U_2$. Along the path abc, $\delta U_2 = 36\text{ J} - \delta W_2$. Equating the two, $30\text{ J} = 36\text{ J} - \delta W_2$, which gives $\delta W_2 = 6\text{ J}$.

   Step Solution:

    1.  Analyze path adc: Use the First Law $\Delta U = Q - W$.

    2.  Calculate $\Delta U$: $\Delta U = 50\text{ J} - 20\text{ J} = 30\text{ J}$.

    3.  Apply State Function property: Note that $\Delta U$ is the same for path abc because initial and final states are identical.

    4.  Analyze path abc: Set up the equation $W_{abc} = Q_{abc} - \Delta U$.

    5.  Final Calculation: $W_{abc} = 36\text{ J} - 30\text{ J} = 6\text{ J}$.

   Difficulty Level: Medium.

   Concept Name: First Law of Thermodynamics / Path Independence of Internal Energy.

   Short cut solution: Since the change in internal energy is constant for both paths, the difference in heat supplied must equal the difference in work done: $Q_1 - W_1 = Q_2 - W_2 \rightarrow 50 - 20 = 36 - W_2 \rightarrow W_2 = 6\text{ J}$.

Question 58

   Question: An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If $Q_1, Q_2, Q_3$ indicate the heat absorbed by the gas along the three processes and $\Delta U_1, \Delta U_2, \Delta U_3$ indicate the change in internal energy along the three processes respectively, then.

   Options: 

    A. $Q_1 > Q_2 > Q_3$ and $\Delta U_1 = \Delta U_2 = \Delta U_3$

    B. $Q_3 > Q_2 > Q_1$ and $\Delta U_1 = \Delta U_2 = \Delta U_3$

    C. $Q_1 = Q_2 = Q_3$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$

    D. $Q_3 > Q_2 > Q_1$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$

   Correct Answer: A.

   Year: 2012 Mains.

   Solution: Change in internal energy is path independent and depends only on the initial and final states; thus, $\Delta U_1 = \Delta U_2 = \Delta U_3$. Work done ($W$) is the area under the P-V graph. From the diagram, the area under curve 1 is greater than 2, which is greater than 3, so $W_1 > W_2 > W_3$. According to the first law ($Q = W + \Delta U$), since $\Delta U$ is constant, $Q_1 > Q_2 > Q_3$.

   Step Solution:

    1.  Evaluate Internal Energy: All three processes connect state A to state B, so the change in internal energy ($\Delta U$) must be equal for all paths.

    2.  Determine Work Done: Identify that work done equals the area under each curve on the P-V diagram.

    3.  Compare Areas: Visually confirm from the graph that Path 1 encloses the most area and Path 3 the least ($W_1 > W_2 > W_3$).

    4.  Relate Heat and Work: Apply the First Law ($Q = \Delta U + W$).

    5.  Conclusion: Because $\Delta U$ is a constant, the heat absorbed $Q$ follows the same ranking as $W$, resulting in $Q_1 > Q_2 > Q_3$.

   Difficulty Level: Easy/Medium.

   Concept Name: Work as Area Under P-V Curve / Internal Energy as a State Function.

   Short cut solution: In P-V diagrams where paths share the same start and end points, $\Delta U$ is always equal; simply rank the heat $Q$ by comparing the area under each curve.

Question 62

   Question: If $\Delta U$ and $\Delta W$ represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

   Options: 

    A. $\Delta U = -\Delta W$, in an adiabatic process

    B. $\Delta U = \Delta W$, in an isothermal process

    C. $\Delta U = \Delta W$, in an adiabatic process

    D. $\Delta U = -\Delta W$, in an isothermal process

   Correct Answer: A.

   Year: 2010.

   Solution: According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. In an adiabatic process, there is no heat exchange, so $\Delta Q = 0$. This results in $0 = \Delta U + \Delta W$, or $\Delta U = -\Delta W$. For an isothermal process, $\Delta U = 0$, so $\Delta Q = \Delta W$, making the other options incorrect.

   Step Solution:

    1.  Define the First Law: Write out $\Delta Q = \Delta U + \Delta W$.

    2.  Apply Process Condition: For an adiabatic process, heat exchange $\Delta Q$ is zero by definition.

    3.  Substitute Zero: $0 = \Delta U + \Delta W$.

    4.  Rearrange the Equation: $\Delta U = -\Delta W$.

    5.  Verify Process Type: Confirm that this specific mathematical relationship characterizes an adiabatic change where work is done at the expense of internal energy.

   Difficulty Level: Easy.

   Concept Name: First Law of Thermodynamics / Adiabatic Process.

   Short cut solution: In an adiabatic process ($\Delta Q = 0$), the system is insulated, so any work done must be exactly balanced by an opposite change in internal energy ($\Delta U + \Delta W = 0$).

Question 65

   Question: The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is.

   Options: 

    A. 6400 J

    B. 5400 J

    C. 7900 J

    D. 8900 J.

   Correct Answer: C.

   Year: 2009.

   Solution: Heat energy given $dQ = dU + dW$ where $dU$ is the change in internal energy and $dW$ is the work done. Given $dQ = 2 \text{ kcal} = 2000 \times 4.2 \text{ J}$ and $dW = 500 \text{ J}$. $\therefore dQ = 2000 \times 4.2 = dU + 500 \Rightarrow dU = 7900 \text{ J}$.

   Step Solution:

    1.  Identify Heat Supplied ($Q$): Convert kcal to Joules. $Q = 2 \text{ kcal} \times 1000 \text{ cal/kcal} \times 4.2 \text{ J/cal} = 8400 \text{ J}$.

    2.  Identify Work Done ($W$): $W = 500 \text{ J}$.

    3.  Apply First Law of Thermodynamics: $\Delta Q = \Delta U + W$.

    4.  Substitute Values: $8400 \text{ J} = \Delta U + 500 \text{ J}$.

    5.  Calculate Internal Energy Change: $\Delta U = 8400 - 500 = 7900 \text{ J}$.

   The difficulty level: Easy.

   The Concept Name: First Law of Thermodynamics / Mechanical Equivalent of Heat.

   Short cut solution: $\Delta U = (2 \times 4200) - 500 = 7900 \text{ J}$.

Question 84

   Question: We consider a thermodynamic system. If $\Delta U$ represents the increase in its internal energy and $W$ the work done by the system, which of the following statements is true?.

   Options: 

    A. $\Delta U = -W$ in an isothermal process

    B. $\Delta U = W$ in an isothermal process

    C. $\Delta U = -W$ in an adiabatic process

    D. $\Delta U = W$ in an adiabatic process.

   Correct Answer: C.

   Year: 1998.

   Solution: According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. In an adiabatic process, no exchange of heat energy takes place, so $\Delta Q = 0$. Thus, $0 = \Delta U + W$, which gives $\Delta U = -W$.

   Step Solution:

    1.  State the First Law: $\Delta Q = \Delta U + W$.

    2.  Define Adiabatic Process: In an adiabatic process, heat exchange $\Delta Q = 0$.

    3.  Substitute $\Delta Q$: $0 = \Delta U + W$.

    4.  Isolate $\Delta U$: Rearrange the equation to find the relationship: $\Delta U = -W$.

    5.  Conclusion: This confirms that work done by the system results in a decrease in internal energy under adiabatic conditions.

   The difficulty level: Easy.

   The Concept Name: First Law of Thermodynamics / Adiabatic Process.

   Short cut solution: In adiabatic changes, $\Delta Q=0$, so $\Delta U$ and $W$ must be equal and opposite ($\Delta U = -W$).

Question 94

   Question: 110 joule of heat is added to a gaseous system whose internal energy is 40J, then the amount of external work done is.

   Options: 

    A. 150J

    B. 70J

    C. 110J

    D. 40J.

   Correct Answer: B.

   Year: 1993.

   Solution: $\Delta Q = \Delta U + \Delta W \Rightarrow \Delta W = \Delta Q - \Delta U = 110 - 40 = 70 \text{ J}$.

   Step Solution:

    1.  Identify Given Values: Heat added $\Delta Q = 110 \text{ J}$ and Internal energy $\Delta U = 40 \text{ J}$.

    2.  Apply First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$.

    3.  Rearrange for Work Done: $\Delta W = \Delta Q - \Delta U$.

    4.  Substitute Values: $\Delta W = 110 \text{ J} - 40 \text{ J}$.

    5.  Final Calculation: $\Delta W = 70 \text{ J}$.

   The difficulty level: Easy.

   The Concept Name: First Law of Thermodynamics.

   Short cut solution: Simply subtract the internal energy change from the heat added: $110 - 40 = 70 \text{ J}$.

Question 101

   Question: A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are $P_A = 3 \times 10^4 \text{ Pa}; V_A = 2 \times 10^{-3} \text{ m}^3; P_B = 8 \times 10^4 \text{ Pa}; V_D = 5 \times 10^{-3} \text{ m}^3$. In the process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be.

   Options: 

    A. 560 J

    B. 800 J

    C. 600 J

    D. 640 J

   Correct Answer: A.

   Year: 1991.

   Solution: Since AB is an isochoric process, no work is done. BC is an isobaric process where $W = P_B \times (V_D - V_A) = 240 \text{ J}$. Therefore, the total heat added is $\Delta Q = 600 + 200 = 800 \text{ J}$. Using the first law, $\Delta Q = \Delta U + \Delta W \Rightarrow \Delta U = \Delta Q - \Delta W = 800 - 240 = 560 \text{ J}$.

   Step Solution:

    1.  Calculate Total Heat ($\Delta Q$): Sum the heat added in both processes: $\Delta Q = Q_{AB} + Q_{BC} = 600 \text{ J} + 200 \text{ J} = 800 \text{ J}$.

    2.  Determine Work for AB ($W_{AB}$): Since the process is isochoric (volume is constant at $V_A$), $W_{AB} = 0$.

    3.  Calculate Work for BC ($W_{BC}$): Use the formula for work at constant pressure: $W_{BC} = P_B(V_D - V_A) = (8 \times 10^4 \text{ Pa}) \times (5 \times 10^{-3} \text{ m}^3 - 2 \times 10^{-3} \text{ m}^3) = 240 \text{ J}$.

    4.  Find Total Work ($\Delta W$): Add the work from both steps: $\Delta W = 0 + 240 \text{ J} = 240 \text{ J}$.

    5.  Apply First Law: Calculate internal energy change using $\Delta U = \Delta Q - \Delta W = 800 \text{ J} - 240 \text{ J} = 560 \text{ J}$.

   Difficulty Level: Hard.

   Concept Name: First Law of Thermodynamics / Isochoric and Isobaric Processes.

   Short cut solution: $\Delta U = (Q_{AB} + Q_{BC}) - [P_B \times (V_D - V_A)] = 800 - 240 = 560 \text{ J}$.

Question 109

   Question: First law of thermodynamics is consequence of conservation of.

   Options: 

    A. work

    B. energy

    C. heat

    D. all of these

   Correct Answer: B.

   Year: 1988.

   Solution: Conservation of energy.

   Step Solution:

    1.  Recall the First Law Equation: Note that $\Delta Q = \Delta U + W$ relates heat, internal energy, and work.

    2.  Identify the Nature of Variables: Recognize that heat ($Q$), internal energy ($U$), and work ($W$) are all different forms of energy.

    3.  Interpret the Law: The law states that the total energy of an isolated system remains constant; it is only transformed from one form to another.

    4.  Relate to Physics Principles: This fundamental principle of energy balance is known as the Law of Conservation of Energy.

    5.  Conclusion: Therefore, the first law is a specific application of the broader principle of energy conservation.

   Difficulty Level: Easy.

   Concept Name: First Law of Thermodynamics / Conservation of Energy.

   Short cut solution: The First Law of Thermodynamics (FLOT) is fundamentally the statement of the Law of Conservation of Energy applied to thermal systems.