Question

An adiabatic piston of mass $m$ equally divides a diathermic container of volume $V_0$, length $l$ and cross-sectional area $A$. A light spring connects the piston to the right wall. In equilibrium, pressure on each side of the piston is $P_0$. The container starts moving with acceleration $a$ towards right. The stretch $x$ in spring is given as $\left(\frac{n}{k+\frac{n{P}_0\gamma A}{l}}\right)$. Find $n$.

$[$ Assume that $x<<l$, the gas in container has the adiabatic exponent $($ ratio of $C_P$ and $C_V)=\gamma ,m=2\text{kg},a=2{\text{m/s}}^2]$

Answer 1

As the container is moving towards the right, from the FBD of the piston, we can say that,

$P_{1}A+kx-P_{2}A=ma........\left(1\right)$

Volume of left portion of container is $V_1=A(\frac{l}{2}-x)$

Volume of right portion of container is $V_2=A(\frac{l}{2}+x)$

Volume of each portions of container in equilibrium, $V_0=A\left(\frac{l}{2}\right)$


Since, the process is adiabatic, using the equation of state of an adiabatic process, we can write that,

$P_0{V}_0^{\gamma }=P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

$\Rightarrow P_1=P_0{\left(\frac{V_0}{V_1}\right)}^{\gamma }=P_0{\left(\frac{l}{l-2x}\right)}^{\gamma }$

$\Rightarrow P_1=P_0{(1-\frac{2x}{l})}^{-\gamma }$

$\Rightarrow P_1=P_0(1+\frac{2\gamma x}{l})(\because x<<l)$

Similarly, $P_2=P_0(1-\frac{2\gamma x}{l})$

Thus, from $\left(1\right)$,

$\left(\frac{4P_0\gamma xA}{l}\right)+kx=ma$

$\Rightarrow x=\frac{ma}{k+\frac{4P_0\gamma A}{l}}$

Substituting the data given in the question, we get,

$x=\frac{4}{k+\frac{4P_0\gamma A}{l}}$

Comparing with the equation given in the question, we get, $n=4$