Question

\(0.5\) mole of an ideal gas at constant temperature \(27^\circ\text{C}\) is kept inside a cylinder of length \(L\) and cross-sectional area \(A\), closed by a massless piston. The cylinder is attached to a conducting rod of length \(L\), cross-sectional area \(\frac{1}{9}\,\text{m}^2\) and thermal conductivity \(k\), whose other end is maintained at \(0^\circ\text{C}\). The piston is moved such that heat flow through the conducting rod is constant. Find the velocity of the piston when it is at a height \(L/2\) from the bottom of the cylinder. (Neglect any loss of heat from the system.) 

• A. \(\dfrac{k}{R}\,\text{m/s}\)
• B. \(\dfrac{k}{10R}\,\text{m/s}\)
• C. \(\dfrac{k}{100R}\,\text{m/s}\)
• D. \(\dfrac{k}{1000R}\,\text{m/s}\)

Hint:

For isothermal processes: \[ dQ = PdV \] Equate heat flow rate with conduction rate for piston motion problems.

Answer 1

The Correct Option is B

Step 1: Heat current through the rod: \[ \frac{dQ}{dt} = \frac{kA(\Delta T)}{L} \]
Step 2: For isothermal process of ideal gas: \[ dQ = PdV \]
Step 3: Using \(PV = nRT\), we get: \[ P = \frac{nRT}{V} \]
Step 4: Since heat flow is constant, equate \(PdV/dt\) to constant heat current and substitute values at \(V=AL/2\).
Step 5: Solving gives piston velocity: \[ v = \frac{k}{10R} \]