Question

0.5 mole of an ideal gas at constant temperature 27°C kept inside a cylinder of length L and cross section area A closed by a massless piston. The cylinder is attached with a conducting rod of length I, cross-section area (1/9) m2 and thermal conductivity k, whose other end is maintained at 0°C. If piston is moved such that rate of heat ow 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)

Answer 1

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$\large \frac{∆Q}{∆t} = \frac {∆W}{∆t}$ = work done per unit time=

$\large \frac {kaθ}{L}$

$\large \frac {dW}{dt} = P \frac { dv } { dt }$ = $\large k \frac {aθ}{L}, P = \frac {nRT}{V}$

$\large → \frac {0.5R(300)}{V} A. \frac {dl}{dt} = \frac {kaθ}{L}$

$\large → \frac {0.5R(300)}{L} A. v = \frac {kaθ}{L}$

$\large → v = \frac {ka}{R} (\frac {27}{300} ) = \frac {k}{100R}$

Answer 2

Answer:

∆Q

=

∆t

∆W

= work done per unit time=

\large \frac {kaθ}{L}

L

kaθ

\large \frac {dW}{dt} = P \frac { dv } { dt }

dt

dW

=P

dt

dv

= \large k \frac {aθ}{L}, P = \frac {nRT}{V}k

L

aθ

,P=

V

nRT

\large → \frac {0.5R(300)}{V} A. \frac {dl}{dt} = \frac {kaθ}{L}→

V

0.5R(300)

A.

dt

dl

=

L

kaθ

\large → \frac {0.5R(300)}{L} A. v = \frac {kaθ}{L}→

L

0.5R(300)

A.v=

L

kaθ

\large → v = \frac {ka}{R} (\frac {27}{300} ) = \frac {k}{100R}→v=

R

ka

(

300

27

)=

100R

k