A piston of mass M$M$ is hung from a massless spring whose restoring force law goes as F = -kx³$F = -kx^3$, where k$k$ is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n$n$' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T$T$) 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 L₀$L_0$ to L₁$L_1$, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)

Related Formula

Core Logic

In a P-V diagram, the work done in a cyclic process is equal to the area of the enclosed loop. Since the cycle ABC is clockwise (A arrow$\rightarrow$ B arrow$\rightarrow$ C arrow$\rightarrow$ A), the net work done by the system is positive.

Step 1: Calculate the Area

The shape ABC is a right-angled triangle. Base of the triangle = Δ V = VC - VA = 5 - 2 = 3 m³$= \Delta V = V_C - V_A = 5 - 2 = 3 \text{ m}^3$ Height of the triangle = Δ P = PB - PC = 300 - 100 = 200 Pa$= \Delta P = P_B - P_C = 300 - 100 = 200 \text{ Pa}$

Step 2: Compute Work Done

Pattern Recognition

Always verify axis units (kPa, atm, Liters, cc) before blindly multiplying area. Here units are cleanly in standard SI (Pa and m³$m^3$), so 1 Pa · 1 m³ = 1 Joule$1 \text{ Pa} \cdot 1 \text{ m}^3 = 1 \text{ Joule}$.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

• First Law of Thermodynamics:

dQ = dU + dW$dQ = dU + dW$

In an adiabatic process:

• Change in Internal Energy:

Core Logic

Let's analyze each statement:

• (A) Internal energy is constant: Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy (U ∝ T$U \propto T$) must change.
• (B) Work done is equal to the change in internal energy: Correct in magnitude (|dW| = |dU|$|dW| = |dU|$). By definition, dW = -dU$dW = -dU$, which correlates the magnitude of work to the change in internal energy.
• (C) Product of temperature and volume is constant: Incorrect. The adiabatic equation of state is T Vγ-1 = constant$T V^{\gamma-1} = \text{constant}$.
• (D) Product of pressure and volume is constant: Incorrect. The relation is P V^γ = constant$P V^\gamma = \text{constant}$.
• (E) Work done is proportional to (T₂ - T₁)$(T_2 - T_1)$: Correct. Since dW = -dU = -n Cv (T₂ - T₁)$dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change (T₂ - T₁)$(T_2 - T_1)$.

Step 1: Determine the correct option

Since only statements (B) and (E) are correct, the correct option is (3).

Pattern Recognition

Sees: Characteristics of adiabatic thermodynamic process. Trap: Confusing adiabatic state relations (PV^γ = C$PV^\gamma = C$, TVγ-1 = C$TV^{\gamma-1} = C$) with isothermal state relations (PV = C$PV = C$, T = C$T = C$). Shortcut: First law of thermodynamics under dQ=0$dQ=0$ strictly enforces |dW| = |dU|$|dW| = |dU|$, which validates statement B and E immediately.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

dQ = n C dT$dQ = n C dT$

Core Logic

An adiabatic process involves no heat exchange between the system and its surroundings (dQ = 0$dQ = 0$).

The molar heat capacity C$C$ is defined as:

Since dQ = 0$dQ = 0$ while the temperature changes (dT ≠ 0$dT \neq 0$):

C = 0$C = 0$

Thus, the molar heat capacity for any adiabatic process is always zero.

Step 1: Check other options

• Option (1): Isothermal processes have infinite molar heat capacity (dT = 0$dT = 0$).
• Option (2): From the First Law (dQ = dU + dW dW = -dU$dQ = dU + dW \implies dW = -dU$), the work done equals the decrease in internal energy.
• Option (4): The internal energy of an ideal gas (dU = n Cv dT$dU = n C_{\mathrm{v}} dT$) increases directly as temperature increases.

Step 2: Final Conclusion

The statement 'The molar heat capacity is zero' is true.

Pattern Recognition

Adiabatic = no heat flow (dQ=0$dQ=0$). Since molar heat capacity tracks the ratio of heat input to temperature change, C$C$ must be 0. Conversely, isothermal has infinite capacity because heat is absorbed without any temperature change (dT=0$dT=0$).

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

For a cyclic thermodynamic process, the net change in internal energy is zero:

By the First Law of Thermodynamics, the total heat exchanged QT$Q_T$ equals the net work done Wₙₑₜ$W_{\text{net}}$:

Core Logic

The cycle consists of three steps:

• Isothermal expansion from (P₀, V₀)$(P_0, V_0)$ to volume 4V₀$4V_0$.
• Isobaric compression to the original volume V₀$V_0$.
• Isochoric heating back to the initial state.

Step 1: Work in Isothermal Expansion (W₁$W_1$)

Initial state: (P₀, V₀)$(P_0, V_0)$. Final state volume: 4V₀$4V_0$.

Also, the pressure at the end of this process is:

Step 2: Work in Isobaric Compression (W₂$W_2$)

The process occurs at constant pressure P = P₁ = P₀/4$P = P_1 = P_0/4$. The volume goes from 4V₀$4V_0$ back to V₀$V_0$:

Step 3: Work in Isochoric Heating (W₃$W_3$)

Since the volume is held constant at V₀$V_0$, no boundary work is done:

W₃ = 0$W_3 = 0$

Step 4: Total Heat Exchanged (QT$Q_T$)

Pattern Recognition

In any cyclic system returning to its initial state, finding total heat is mathematically equivalent to calculating the enclosed area on a P-V diagram. Here, the isobaric step occurs at the lowest expanded pressure, resulting in a simple negative rectangular area correction subtracted from the logarithmic isothermal expansion curve.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

The work done W$W$ in a cyclic thermodynamic process is equal to the area enclosed by the loop on a Pressure-Volume (P-V$P-V$) diagram:

For an ellipse with semi-major axis a$a$ and semi-minor axis b$b$:

Core Logic

Determine the semi-axes of the elliptical cycle on the P-V$P-V$ plane:

• On the Pressure axis (x-axis):

• On the Volume axis (y-axis):

Step 1: Calculate Area

Substitute a$a$ and b$b$ in standard SI units into the area equation:

Express in terms of × 10⁻¹ ~J$\times 10^{-1} \mathrm{~J}$:

Thus, the multiplier is 314$314$.

Pattern Recognition

Sees: Circular/elliptical thermodynamic cycle. Shortcut: Work done is always π Δ P Δ V / 4$\pi \Delta P \Delta V / 4$. Simply compute the semi-axes difference 100 ~kPa$100 \mathrm{~kPa}$ and 100 ~cm³$100 \mathrm{~cm^3}$ and multiply by π$\pi$ directly, keeping tracking of metric prefixes (10³ × 10⁻⁶ = 10⁻³$10^3 \times 10^{-6} = 10^{-3}$).

Chapter Mix

Class 11 Physics: Thermodynamics