0.5 mole of an ideal gas at constant temperature 27^ ^ C kept...

0.5 mole of an ideal gas at constant temperature

27^{{}^{\circ}} \mathrm{C}

kept inside a cylinder of length

L

and crosssection area A closed by a massless piston. 0.5 mole of an ideal gas at constant temperature 27^ ^ C kept inside a cylinder of length L and crosssection area A

0.5 mole of an ideal gas at constant temperature 27^ ^ C kept inside a cylinder of length L and crosssection area A

The cylinder is attached with a conducting rod of length

L,

cross-section area

(1 / 9) \mathrmm^{2}

and thermal conductivity

\mathrm{k}

, whose other end is maintained at

0^{{}^{\circ}} \mathrm{C}

. If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height

L / 2

from the bottom of cylinder is : [Neglect any kind of heat loss from system

Options

A $\left(\frac{\mathrm{k}}{R}\right) \mathrm{m} / \mathrm{sec}$
B $\left(\frac{\mathrm{k}}{10 R}\right) \mathrm{m} / \mathrm{sec}$
C $\left(\frac{\mathrm{k}}{100 R}\right) \mathrm{m} / \mathrm{sec}$
D $\left(\frac{\mathrm{k}}{1000 R}\right) \mathrm{m} / \mathrm{sec}$

Correct answer

C. $\left(\frac{\mathrm{k}}{100 R}\right) \mathrm{m} / \mathrm{sec}$

Step-by-step solution

\frac{\Delta Q}{\Delta t}=\frac{\Delta W}{\Delta t}=

work done per unit time =

\frac{\mathrm{k} a \theta}{L}\frac{d W}{d t}=

\frac{d v}{d t}=\mathrm{k} \frac{a \theta}{L},

\frac{n R T}{V}\Rightarrow \quad \frac{0.5 R(300)}{V} A \cdot \frac{d \ell}{d t}=\frac{\mathrm{k} a \theta}{L} ⇒ \frac{0.5 R(300)}{A \cdot \frac{L}{2}}

\cdot

\frac{\mathrm{k} a \theta}{L}\Rightarrow \quad v=\frac{\mathrm{k} a}{R}\left(\frac{27}{300}\right)=\frac{\mathrm{k}}{100 R}

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