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

where, Δ U$\Delta U$ = change in internal energy, Δ Q$\Delta Q$ = heat exchange, Δ Won system$\Delta W_{\text{on system}}$ = work done on the system.

Core Logic

During the melting of ice at 273~K$273\mathrm{~K}$, the density of water is higher than the density of ice. This means the volume of the ice-water system decreases during melting:

Since the system contracts, the atmosphere performs positive work on it:

Additionally, heat is absorbed by the system to melt the ice, so Δ Q > 0$\Delta Q > 0$. By the first law of thermodynamics, since both Δ Q$\Delta Q$ and Δ Won system$\Delta W_{\text{on system}}$ are positive, the internal energy of the system increases:

Step 1: Evaluation of Options

Let's check the given options:

• Internal energy remains unchanged arrow$\rightarrow$ False (it increases).
• Positive work is done by the system arrow$\rightarrow$ False (work done by the system is negative since it contracts).
• Internal energy decreases arrow$\rightarrow$ False.
• Positive work is done on the ice-water system by the atmosphere arrow$\rightarrow$ True (since volume decreases under atmospheric pressure).

Pattern Recognition

Remember: Ice contracts upon melting (unlike most solids). Shrinking volume (V$V \downarrow$) under positive pressure (P > 0$P > 0$) means the surroundings (atmosphere) compress it, performing positive work on the system.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter Class 11 Physics: Thermodynamics

Related Formula

For an adiabatic process:

where, T$T$ = absolute temperature in Kelvin, V$V$ = volume of the gas, γ$\gamma$ = adiabatic exponent.

Core Logic

Given values:

• Initial volume, V₁ = 800~cm³$V_1 = 800\mathrm{~cm}^3$
• Final volume, V₂ = 200~cm³$V_2 = 200\mathrm{~cm}^3$
• Initial temperature, T₁ = 27°C = 27 + 273 = 300~K$T_1 = 27^{\circ}\mathrm{C} = 27 + 273 = 300\mathrm{~K}$
• Adiabatic exponent, γ = 1.5 γ - 1 = 0.5$\gamma = 1.5 \implies \gamma - 1 = 0.5$

Step 1: Calculating Final Temperature

Apply the adiabatic relation:

Substitute the values:

Step 2: Calculating Change in Temperature

Now compute the change in temperature (Δ T$\Delta T$):

Pattern Recognition

Always read carefully to see if the question asks for the final temperature or the change in temperature. Many students lose marks by choosing 600~K$600\mathrm{~K}$ (the final temperature) instead of the difference 300~K$300\mathrm{~K}$! Stay sharp.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

First Law of Thermodynamics:

Core Logic

• Isobaric: Pressure is constant, work done Δ W = PΔ V$\Delta W = P\Delta V$. Thus, Δ Q = Δ U + PΔ V$\Delta Q = \Delta U + P\Delta V$ (Matches IV).
• Isochoric: Volume is constant, Δ V = 0 Δ W = 0$\Delta V = 0 \implies \Delta W = 0$. Thus, Δ Q = Δ U$\Delta Q = \Delta U$ (Matches II).
• Adiabatic: No heat transfer, Δ Q = 0$\Delta Q = 0$ (Matches III).
• Isothermal: Temperature is constant, internal energy change Δ U = 0$\Delta U = 0$ for an ideal gas. Thus, Δ Q = Δ W$\Delta Q = \Delta W$ (Matches I).

Step 1: Match Compilation

Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I).

Pattern Recognition

Adiabatic definition is always heat-isolated (Q=0$Q=0$), isochoric implies rigid boundaries (W=0$W=0$). Identifying these two immediately isolates the correct match permutation in seconds.

Chapter Mix

Class 11 Physics: Thermodynamics

Core Logic

Let's evaluate each process condition based on the first law of thermodynamics:

• (A) Isothermal: Continuous constant temperature (Δ T = 0$\Delta T = 0$) implies that the internal energy change of an ideal gas is zero, so Δ U = 0 (IV)$\Delta U = 0 \implies \text{(IV)}$ [cite: 730].
• (B) Adiabatic: No thermal energy transfer occurs between the system and surroundings, meaning Δ Q = 0 (II)$\Delta Q = 0 \implies \text{(II)}$ [cite: 731].
• (C) Isobaric: Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Δ U ≠ 0 (III)$\Delta U \neq 0 \implies \text{(III)}$ [cite: 732].
• (D) Isochoric: Rigid boundary condition at constant volume (Δ V = 0$\Delta V = 0$) ensures work done Δ W = PΔ V = 0 (I)$\Delta W = P\Delta V = 0 \implies \text{(I)}$ [cite: 733].

Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729].

Pattern Recognition

Matching 'Isochoric' with zero work done (Δ W=0$\Delta W=0$) or 'Adiabatic' with zero heat exchange (Δ Q=0$\Delta Q=0$) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733].

Chapter Mix

Class 11 Physics: Thermodynamics