Question

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross sectional area $4\times {10}^{-3}{m}^2$. Initially the gas is at ${27}^{\circ }C$ and volume of gas is $1.2\times {10}^{-3}{m}^3$, system is in equilibrium and spring is in natural state. The sample is heated by a heater until piston moves out slowly by $5\text{cm}$. Calculate final temperature of the gas and heat supplied by heater. Take $k$ (spring constant) = $8000N/m$, mass of piston = $20\text{kg}$ (Assume there is no loss of heat through coils and walls, take $P_{a}tm={10}^5\text{Pa}$ )

Final temperature of the gas (in $K$):
Heat supplied by heater (in $J$):

Answer 1

The piston is moving slowly $\Rightarrow$ quasistatic process
Let us assume that at any given time the piston has moved by a distance $x$.
$x_i=0$
It is given that the piston is always in the equlibrium $\Rightarrow$ the net force on piston is zero at all times.
Considering the FBD of the piston,
$P_{gas}A=P_{o}A+mg+kx$
$P_{gas}=P_o+\frac{mg}{A}+\frac{kx}{A}$
Since the is no addition or loss of the gas, the number of moles of the ideal gas in the cylinder remains conserved.
$\therefore n_i=n_f$
Using ideal gas equation,
$\frac{P_i{V}_i}{T_i}=\frac{P_f{V}_f}{T_f}$
Substituting the values of pressures and volumes with corresponding value of $x$,we have
$\frac{(P_o+\frac{mg}{A})V_o}{T_o}=\frac{(P_o+\frac{mg}{A}+\frac{k{x}_f}{A})(V_o+A{x}_f)}{T_o}$
Substituting all the values we get, $T_f=583.33K\simeq 583K$

In the second part we need to find the heat supplied to the system,
$\Delta Q=\Delta U+W$
where, $\Delta U=n{C}_v\Delta T$...(i)
Since the number of moles is not speciefied in the problem, we make use of the ideal gas equation,
$PV=nRT\Rightarrow \Delta \left(PV\right)=nR\Delta T$
$\Rightarrow n\Delta T=\frac{P_f{V}_f-P_i{V}_i}{R}$
For monoatomic gas, $C_v=\frac{3}{2}R$
Putting these in equation (i)
$\Delta U=\frac{3}{2}R\frac{P_f{V}_f-P_i{V}_i}{R}=255J$
Workdone by the gas $W_{gas}=\int PdV={\int }_{x=0}^{x=5}(P_o+\frac{mg}{A}+\frac{kx}{A})Adx$
Integrating ans substing we get, $W=40J$

The gas is doing work on the piston. The piston is moving slowly $\Rightarrow$ KE is not changing.
Using Work-Energy theorem,
$W_{net}$ on the piston =$\Delta KE=0$
Again using the FBD of piston,
$W_{gravity}+W_{spring}+W_{atm}+W_{gas}=0$
$W_{gravity}=-mg{x}_f,W_{spring}=-\frac{1}{2}k{x}_f^2,$ $W_{atm}=-P_{o}A{x}_f$
Substituting these in the above equation,
$W_{gas}=mg{x}_f+\frac{1}{2}k{x}_f^2+P_{o}A{x}_f$
Putting the values,
$W_{gas}=20\times 100\times 0.05+\frac{1}{2}8000\times (0.05{)}^2+{10}^5\times 4\times {10}^{-3}0.05$
$W_{gas}=40J$

$\Delta Q=\Delta U+W=255+40=295J$