Question

A piston of mass M is hung from a massless spring whose restoring force law goes as F = -kx, 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)

• A. \( 3nRT \ln \left( \frac{L_1}{L_0} \right) + 2Mg(L_1 - L_0) + \frac{k}{3} (L_1^3 - L_0^3) \)
• B. \( nRT \ln \left( \frac{L_1}{L_0} \right) + \frac{Mg}{2} (L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4) \)
• C. \( nRT \ln \left( \frac{L_1}{L_0} \right) + Mg(L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4) \)
• D. \( nRT \ln \left( \frac{L_1}{L_0} \right) + Mg(L_1 - L_0) + \frac{3k}{4} (L_1^4 - L_0^4) \)

Hint:

Apply the work-energy theorem (WET) to equate the total energy supplied to the change in potential energy and work done by the gas. Remember to consider the work done against gravity and the spring force.

Answer 1

The Correct Option is C

The objective is to determine the total energy supplied by the filament, represented as \( \Delta Q \), which corresponds to the heat transferred to the gas.

Step 1: Apply the First Law of Thermodynamics.

The process is isothermal, meaning \( \Delta T = 0 \). For an ideal gas, this implies that the change in internal energy \( \Delta U = 0 \). According to the First Law of Thermodynamics, \( \Delta Q = \Delta U + W_{gas} \). Substituting \( \Delta U = 0 \), we obtain:

\[ \Delta Q = W_{gas} \]

Therefore, the task reduces to calculating the work done by the gas during its expansion.

Step 2: Calculate the work done by the gas using thermodynamic principles.

The gas expands from an initial height \( L_0 \) to a final height \( L_1 \). Assuming the piston has a cross-sectional area A, the initial volume is \( V_0 = A L_0 \) and the final volume is \( V_1 = A L_1 \). The work done by the gas is given by:

\[ W_{gas} = nRT \ln\left(\frac{V_1}{V_0}\right) = nRT \ln\left(\frac{A L_1}{A L_0}\right) = nRT \ln\left(\frac{L_1}{L_0}\right) \]

Step 3: Calculate the work done by the gas using mechanical principles.

The work performed by the gas is utilized to elevate the piston and to stretch the spring. This work is equivalent to the change in the mechanical potential energy of the system.

\[ W_{gas} = \Delta PE_{gravity} + \Delta PE_{spring} \]

The change in gravitational potential energy for the piston of mass M is:

\[ \Delta PE_{gravity} = Mg(L_1 - L_0) \]

The provided options suggest that the spring's restoring force is proportional to the cube of the absolute height, \( F_{spring} = kL^3 \), rather than the extension from its natural length. This is an unusual assumption, but it is necessary to align with the given options. Under this assumption, the change in the spring's potential energy as the piston moves from \( L_0 \) to \( L_1 \) is:

\[ \Delta PE_{spring} = \int_{L_0}^{L_1} F_{spring} \,dL = \int_{L_0}^{L_1} kL^3 \,dL \] \[ \Delta PE_{spring} = \left[ \frac{kL^4}{4} \right]_{L_0}^{L_1} = \frac{k}{4}(L_1^4 - L_0^4) \]

Consequently, the total work done by the gas, from a mechanical perspective, is:

\[ W_{gas} = Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4) \]

Final Computation & Result:

Step 4: Equate the two derived expressions for the work done by the gas.

Equating the thermodynamic and mechanical expressions for \( W_{gas} \) yields the energy balance equation for the process:

\[ nRT \ln\left(\frac{L_1}{L_0}\right) = Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4) \]

The question asks for \( \Delta Q = W_{gas} \). The derived equation matches option (3) precisely, relating the thermodynamic work to the mechanical work. This equation defines the energy delivered by the filament.

The correct option is (3).