A piston of mass M is hung from a massless spring whose restoring force law goes as F = -kx^3, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L_0 to L_1, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
Piston connected to spring with gas underneath for Q11
A schematic of a piston of mass M connected to a spring inside a vertical chamber, separating gas at the bottom from vacuum/atmosphere at the top.

Solution & Explanation

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + W_textby gas Work done by an ideal gas during isothermal expansion: W_textgas = nRTlnleft(fracV_1V_0right) = nRTlnleft(fracL_1L_0 ight) Conservation of Energy (Work-Energy Theorem): Total energy delivered by the heating filament (W_textfilament) must equal the total work needed to lift the piston against gravity and compress the non-linear spring. ### Core Logic Since the process is isothermal, the change in internal energy of the ideal gas is zero (Delta U = 0). Hence: Q = W_textgas By the Work-Energy Theorem for the piston: W_textgas + W_textfilament = Delta U_textgravity + Delta U_textspring Let's evaluate each term: - Increase in gravitational potential energy: Delta U_textgravity = Mg(L_1 - L_0) - Increase in spring potential energy: U_textspring = -int_L_0^L_1 F_textrestoring dx = int_L_0^L_1 kx^3 dx = frack4(L_1^4 - L_0^4) ### Step 1: Finding Total Energy Delivered Isolating W_textfilament (the net external energy delivered to the gas system): W_textfilament = W_textgas + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4) Since W_textgas = nRTlnleft(fracL_1L_0 ight): W_textfilament = nRTlnleft(fracL_1L_0 ight) + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4) ### Pattern Recognition Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity (Mgh), and the potential energy of the spring (integrated from kx^3). Keeping this total energy ledger in mind prevents tedious mathematical tangents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics: First Law Class 11 Physics: Work, Energy and Power: Variable Force Integration

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 2

Q5 jee_main_2025_08_april_evening Specific Heat Capacity
Water falls from a height of 200mathrm~m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take g = 10mathrm~m/s^2, specific heat of water = 4200mathrm~J/(kgcdot K))
  • A. 0.23mathrm~K
  • B. 0.36mathrm~K
  • C. 0.14mathrm~K
  • D. 0.48mathrm~K

Solution

### Related Formula Delta U = mgh quad textand quad Q = msDelta T By conservation of energy (assuming all potential energy goes into heating the water): mgh = msDelta T implies Delta T = fracghs where, g = acceleration due to gravity h = height of the fall s = specific heat of water Delta T = rise in temperature ### Core Logic Given parameters: - h = 200mathrm~m - g = 10mathrm~m/s^2 - s = 4200mathrm~J/(kgcdot K) Substitute the values to find Delta T: Delta T = frac10 times 2004200 = frac20004200 Delta T = frac1021 approx 0.476mathrm~K approx 0.48mathrm~K ### Pattern Recognition Sees: "Water falling from height h raises temperature" → Mass cancels out. Delta T = fracghs. Shortcut: Always use SI units (s_textwater = 4200mathrm~J/kgcdot K is given; if given in mathrmcal/gcdot^circ C, convert using 1mathrm~cal = 4.184mathrm~J). ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Work, Energy and Power
Q11 jee_main_2025_08_april_evening Thermodynamic Processes
A monoatomic gas having gamma = frac53 is stored in a thermally insulated container and the gas is suddenly compressed to left(frac18right)^mathrmth of its initial volume. The ratio of final pressure and initial pressure is: (gamma is the ratio of specific heats of the gas at constant pressure and at constant volume)
  • A. 16
  • B. 40
  • C. 32
  • D. 28

Solution

### Related Formula P_i V_i^gamma = P_f V_f^gamma where, P_i, P_f = initial and final pressures V_i, V_f = initial and final volumes gamma = adiabatic exponent ### Core Logic Since the gas is stored in a "thermally insulated container" and is compressed "suddenly", the process is **adiabatic**. From the adiabatic relation: fracP_fP_i = left(fracV_iV_fright)^gamma Given: - V_f = frac18 V_i implies fracV_iV_f = 8 - gamma = frac53 ### Step 1: Computation Substitute the values to find the pressure ratio: fracP_fP_i = (8)^5/3 = left(2^3right)^5/3 fracP_fP_i = 2^5 = 32 ### Pattern Recognition Sees: "suddenly compressed" or "thermally insulated container" → Adiabatic process. Shortcut: P V^gamma = textconstant. Since the volume goes down by 8 times, the pressure increases by 8^gamma = 8^5/3 = 32 times. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q jee_main_2025_29_jan_evening Isothermal and Adiabatic Processes
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

**Assertion** (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Reason (R): In isothermal process, PV = textconstant, while in adiabatic process PV^gamma = textconstant. Here gamma is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below:
  • A. textBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)
  • B. text(A) is true but (R) is false
  • C. textBoth (A) and (R) are true and (R) is the correct explanation of (A)
  • D. text(A) is false but (R) is true

Solution

### Related Formula left(fracdPdVright)_textisothermal = -fracPV left(fracdPdVright)_textadiabatic = -gamma fracPV ### Core Logic The slope of an adiabatic process on a P-V diagram is gamma times steeper than that of an isothermal process: left|left(fracdPdVright)_textadiabaticright| > left|left(fracdPdVright)_textisothermalright|
Isothermal and Adiabatic Processes diagram for Q2 - JEE Main 2025 Evening
Isothermal and Adiabatic Processes diagram for Q2 - JEE Main 2025 Evening
When pressure increases (compression), the volume drops. Because the adiabatic curve is steeper, the pressure rises more rapidly for a given drop in volume, or conversely, for a specified increase in pressure, the volume falls off more rapidly in the isothermal process than the adiabatic process. Hence, Assertion (A) is true. Reason (R) states the governing equations PV = C and PV^gamma = C, which directly lead to these slope expressions via differentiation. Thus, Reason (R) is true and correctly explains Assertion (A). ### Pattern Recognition Adiabatic curves are steeper than isothermal curves on a P-V diagram because gamma > 1. For any expansion or compression process, remember that slope magnitude satisfies textSlope_textadi = gamma cdot textSlope_textiso. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q6 jee_main_2025_29_jan_evening Heat and Work in Thermodynamic Processes
A poly-atomic molecule (C_V = 3R, C_P = 4R, where R is gas constant) goes from phase space point A(P_A = 10^5mathrm~Pa, V_A = 4 times 10^-6mathrm~m^3) to point B(P_B = 5 times 10^4mathrm~Pa, V_B = 6 times 10^-6mathrm~m^3) to point C(P_C = 10^4mathrm~Pa, V_C = 8 times 10^-6mathrm~m^3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:
Heat and Work in Thermodynamic Processes diagram for Q6 - JEE Main 2025 Evening
The graph depicts a pressure vs volume plot indicating paths from state A to B (adiabatic) and from B to C (isothermal).
  • A. 500 mathrm~R(ln 3 + ln 4)
  • B. 450 mathrm~R(ln 4 - ln 3)
  • C. 500 mathrm~Rln 2
  • D. 400 mathrm~R ln 4

Solution

### Related Formula Delta Q_textnet = Delta Q_AB + Delta Q_BC Delta Q_textisothermal = nRT lnleft(fracV_fV_i ight) = P_i V_i lnleft(fracV_fV_i ight) ### Core Logic For path A rightarrow B: Since it is given as an adiabatic path: Delta Q_AB = 0 For path B rightarrow C: Since it is given as an isothermal path, the change in internal energy Delta U_BC = 0. From the first law of thermodynamics, heat absorbed equals work done: Delta Q_BC = W_BC = nRT_B lnleft(fracV_CV_B ight) Using the ideal gas state at point B, nRT_B = P_B V_B: P_B V_B = (5 times 10^4 mathrm~Pa) times (6 times 10^-6 mathrm~m^3) = 0.3 mathrm~J Wait, let's express it in terms of the gas constant R for a single mole (n=1) using temperature data directly provided in the original figure labels (T_B = 450mathrm~K): Delta Q_BC = (1) cdot R cdot (450) cdot lnleft(frac8 times 10^-66 times 10^-6right) Delta Q_BC = 450 R lnleft(frac43 ight) = 450 R (ln 4 - ln 3) Thus, the total net heat absorbed per unit mole is: Delta Q = 0 + 450 R (ln 4 - ln 3) = 450 R (ln 4 - ln 3) ### Pattern Recognition Adiabatic paths have zero heat exchange by baseline definition. The calculation boils down directly to the work done during the isothermal stage B rightarrow C matching RT ln(V_f/V_i). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q7 jee_main_2025_28_jan_morning Carnot Engine and Efficiency
A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine mathrmE_1 works between 473K to 373K and engine mathrmE_2 works between 373K to 273K. If eta_12 , eta_1 and eta_2 are the efficiencies of the engines E, mathrmE_1 and mathrmE_2 , respectively, then
  • A. eta_12 < eta_1 + eta_2
  • B. eta_12 = eta_1eta_2
  • C. eta_12 = eta_1 + eta_2
  • D. eta_12 geq eta_1 + eta_2

Solution

### Related Formula eta = 1 - fracmathrmT_LmathrmT_H ### Core Logic Let's compute the efficiency parameters explicitly: eta_12 = 1 - frac273473 = frac200473 approx 0.423 eta_1 = 1 - frac373473 = frac100473 approx 0.211 eta_2 = 1 - frac273373 = frac100373 approx 0.268 Evaluating the linear sum of fractional bounds: eta_1 + eta_2 = 0.211 + 0.268 = 0.479 Comparing the outputs clearly demonstrates: eta_12 < eta_1 + eta_2 ### Step 1: Final Conclusion Thus, the inequality satisfies option (1). ### Pattern Recognition The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-eta_12) = (1-eta_1)(1-eta_2), which algebraically forces eta_12 = eta_1 + eta_2 - eta_1eta_2 < eta_1 + eta_2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

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