Ideal gas at pressure $10^{5} Pa$ and volume $1 m^{3}$ is enclosed by a piston in a cylinder. We start to move the piston outwards at a constant velocity of 1 cm/s. The cross–sectional area of the piston is $0.1 m^{2}$ . While the piston is moving, we can deliver heat to the gas through a heating filament.
How should the heating power change as a function of time if we keep the temperature of the gas constant ? (Apart from the heat transfer between the gas and the heating filament all other heat exchange can be neglected).

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$P = \frac{100}{1 + 10^{- 3} t}$

$P = 100 + 10^{- 3} t$

$P = \frac{100}{1 + 10^{- 3} t^{2}}$

None of these

answer is A.

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Detailed Solution

where in our case $p V = p_{0} V_{0}$ , $P = \frac{10^{5} \frac{N}{m^{2}} .1 m^{3} .0.1 m^{2} {.10}^{- 2} \frac{m}{s}}{1 m^{3} + 0.1 m^{2} {.10}^{- 2} \frac{m}{s} . t} = \frac{100}{1 + 10^{- 3} \frac{1}{s} . t} W .$
and so the pressure expressed with the variables of the process is $p = \frac{p_{0} V_{0}}{V_{0} + A v t}$
and with it the power as a function of time is $P = \frac{p_{0} V_{0} A v}{V_{0} + A v t}$
Numerically,

$P = \frac{10^{5} \frac{N}{m^{2}} .1 m^{3} .0.1 m^{2} {.10}^{- 2} \frac{m}{s}}{1 m^{3} + 0.1 m^{2} {.10}^{- 2} \frac{m}{s} . t} = \frac{100}{1 + 10^{- 3} \frac{1}{s} . t} W .$