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

For a diatomic gas, γ = 1.4$\gamma = 1.4$. Given initial temperature T = 27°C = 300 K$T = 27^{\circ}\mathrm{C} = 300 \, \mathrm{K}$. Since Wisothermal = Wadiabatic$W_{\text{isothermal}} = W_{\text{adiabatic}}$, we equate the two expressions.

Step 1: Compute Work Done in Isothermal Expansion

Step 2: Equate and Solve for Final Temperature

Equating (1) to adiabatic work:

Step 3: Convert to Celsius

Closest option is -56$-56$.

Pattern Recognition

Equating the work of two different processes links the log expansion to the temperature drop. Recognize diatomic gas instantly means γ = 1.4$\gamma = 1.4$, providing the 0.4$0.4$ divisor.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

W = P Δ V$W = P \Delta V$

Core Logic

First, calculate the heat supplied Δ Q$\Delta Q$:

Next, calculate the work done W$W$ against atmospheric pressure due to volume change: W = P Δ V$W = P \Delta V$

Given P = 10⁵ Pa$P = 10^5 \text{ Pa}$:

Step 1: Internal Energy Calculation

Using the First Law of Thermodynamics:

Pattern Recognition

When heating solids or liquids, the expansion is so small that Δ V$\Delta V$ is negligible, meaning W ≈ 0$W \approx 0$, making Δ U ≈ Δ Q$\Delta U \approx \Delta Q$. Subtracting the exact 0.8J$0.8\text{J}$ provides the highly precise answer required here.

Chapter Mix

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

Related Formula

From Ideal Gas Law for Isochoric Process:

Core Logic

From the P-V graph, the process from P₁$P_1$ to P₂$P_2$ takes place at a constant volume V = 1 m³$V = 1 \text{ m}^3$.

Step 1: Calculate Heat Involved

Multiply and divide by R$R$:

Step 2: Substitution

Given Cv = 21 J/K· mol$C_v = 21 \text{ J/K\cdot mol}$, R = 8.3 J/mol· K$R = 8.3 \text{ J/mol\cdot K}$, P₂ = 30$P_2 = 30$, P₁ = 21.7$P_1 = 21.7$, and V = 1$V = 1$:

Pattern Recognition

For constant volume processes, Δ Q$\Delta Q$ strictly scales with Δ P$\Delta P$. Re-writing n Δ T$n \Delta T$ as V Δ P / R$V \Delta P / R$ circumvents calculating absolute temperatures entirely.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

Cv = CP - R$C_v = C_P - R$

Core Logic

For an isochoric process (constant volume), the heat given is strictly equal to internal energy change:

Step 1: Find moles (n)

Step 2: Evaluating the output

Our Ans. (Bonus) We cannot definitively find the mass (m$m$) in grams because the molar mass (M$M$) of the specific ideal gas is not given in the problem statement. (Note: NTA official answer is 481. This implies assuming M ≈ 1000$M \approx 1000$ or an error in problem specification).

Pattern Recognition

Always extract Cv$C_v$ from Cₚ$C_p$ using Mayer's relation for constant volume heat transfers. If an exam sets a numerical without giving molar mass, it relies on partial contextual hints or is marked as bonus.

Chapter Mix

Class 11 Physics: Thermodynamics

Related Formula

Core Logic

The total work done in a cyclic process is the enclosed area. The cycle is clockwise, but wait—the arrows in the diagram point anti-clockwise (A to B to C and back to A via horizontal top line). Anti-clockwise cycle means negative work. The enclosed area is bounded by a parabola.

Step 1: Identifying the Parabola Parameters

Equation of curve: (V - 2)² = 4ap$(V - 2)^2 = 4ap$ This is an upward-opening parabola with vertex at V = 2$V = 2$, p = 0$p = 0$ (assuming a>0$a>0$). Let's check points A (V=1$V=1$) and C (V=3$V=3$). At V=1$V=1$: (1 - 2)² = 4ap ⇒ 1 = 4ap₀ ⇒ p₀ = (1)/(4a)$(1 - 2)^2 = 4ap \Rightarrow 1 = 4ap_0 \Rightarrow p_0 = \frac{1}{4a}$ At V=3$V=3$: (3 - 2)² = 4ap ⇒ 1 = 4ap₀ ⇒ p₀ = (1)/(4a)$(3 - 2)^2 = 4ap \Rightarrow 1 = 4ap_0 \Rightarrow p_0 = \frac{1}{4a}$ So the top horizontal line is at p₀ = (1)/(4a)$p_0 = \frac{1}{4a}$.

Step 2: Area Calculation

The enclosed shape is bounded by the line p = p₀$p = p_0$ and the parabola p = ((V - 2)²)/(4a)$p = \frac{(V - 2)^2}{4a}$. The area of a parabolic segment (bounded by a chord) is (2)/(3) × Area of enclosing rectangle$\frac{2}{3} \times \text{Area of enclosing rectangle}$. Rectangle width = Δ V = 3 - 1 = 2$\Delta V = 3 - 1 = 2$. Rectangle height = p₀ - 0 = p₀$p_0 - 0 = p_0$. Area = (2)/(3) × (p₀) × (2) = (4p₀)/(3)$\frac{2}{3} \times (p_0) \times (2) = \frac{4p_0}{3}$.

Step 3: Substitute and Determine Sign

Substitute p₀ = (1)/(4a)$p_0 = \frac{1}{4a}$:

The cycle is counter-clockwise, so the net work done by the gas is negative.

Pattern Recognition

For parabolic curves y = kx²$y = kx^2$, the area inside the 'bowl' up to a chord is 2/3$2/3$ of the bounding rectangle. Direction of arrows dictates the sign: clockwise = positive, anti-clockwise = negative.

Chapter Mix

Class 11 Physics: Thermodynamics