Question

A vertical cylinder piston system has cross-section $S$. It contains $1$ mole of an ideal monatomic gas under a piston of mass $M$. At a certain instant a heater is switched on which transmits heat $q$ per unit time of the cylinder. Assume the piston is in equilibrium at all times during the process. The velocity $v$ of the piston given the condition that pressure under the piston is constant and the system is thermally insulated is $\frac{xq}{5S(P_o+\frac{Mg}{S})}$. Here the value of $x$ is

Answer 1

Since the piston is in equilibrium at all times the process is quasi static.
Considering the forces acting on the piston,
$P_{gas}S=P_{o}S+Mg$
$P_{gas}=P_o+\frac{Mg}{S}\Rightarrow$ constant
Hence it is an isobaric process.
If heat is provided to this system both volume and temperature increases.
Rate of heat released to the system $\frac{dQ}{dt}=qJ/s$

Using first law of thermodynamics,
$\Delta Q=\Delta U+W$
Writing this in differential form w.r.t time,
$\frac{dQ}{dt}=\frac{dU}{dt}+\frac{dW}{dt}$
$q=n{C}_v\frac{dT}{dt}+\frac{dW}{dt}$

Since the piston is in equilibrium at all times, the velocity with which the piston moves is constant$\Rightarrow$ KE is constant.
$\therefore W=PdV$

Using this,
$q=n{C}_v\frac{dT}{dt}+\frac{dPdV}{dt}$...(i)
Using ideal gas equation,
$P\frac{dV}{dt}=nR\frac{dT}{dt}$...(ii)
For monoatomic gas, $C_v=\frac{3}{2}R$
Let at some time $t$ the piston be at a distance $x$ from the initial position and it moves further by a distance $dx$ in time $dt$
At this time, change in volume, $\Delta V=Sx$and rate of change of volume, $\frac{dV}{dt}=S\frac{dx}{dt}=Sv$
So we convert all the terms involving temperature into volume in equation (i) using equation (ii)
$q=\frac{3}{2}P\frac{dV}{dt}+P\frac{dV}{dt}=\frac{5}{2}P\frac{dV}{dt}=\frac{5}{2}PSv$
$v=\frac{2q}{5PS}$
$v=\frac{2q}{5S(P_o+\frac{Mg}{S})}$