Question

A heat-conducting piston can move freely inside a closed thermally isolated cylinder which contains an ideal gas. In equilibrium the piston divides the cylinder in two equal parts, the temperature of the gas being $T_{0}$ . The piston is slowly displaced. The adiabatic exponent of the gas is $\gamma =2$ . The temperature of the gas when the volume of one part is $n\left(n = 2\right)$ times than that of the other part is given by $\frac{3}{\alpha \sqrt{\beta }}T_{0}$ . The value of $\left(\alpha + \beta \right)$ is.

Answer 1

Let $d$ be the distance through which the piston has to be moved to make the volume of one part $n$ times that of the other part. Then
$n=\frac{V_{0} + Ad}{V_{0} - Ad}$ or $d=\frac{n - 1}{n + 1}\times \frac{V_{0}}{A}$
Now consider any displacement of the piston by $x$ . since the piston is conducting and the process of compression is slow, the temperature of gases on the two sides of the piston will be the same.
$\therefore \frac{p_{0} V_{0}}{T_{0}}=\frac{p_{1} \left(V_{0} + Ax\right)}{T}=\frac{p_{2} \left(V_{0} - Ax\right)}{T}$
Since it is a quasi- static process
$p_{1}A+F_{agent}=p_{2}A$ or $F_{agent}=\left(p_{2} - p_{1}\right)A$
Then, $dW$ , elementary work done by agent
$=F_{agent}dx=\left(p_{2} - p_{1}\right)A \, dx$
or $dW=\frac{m}{M}RT\left(\frac{1}{V_{0} - Ax} - \frac{1}{V_{0} + Ax}\right)Adx$
$=\frac{m}{M}RT\frac{2 A^{2} xdx}{V_{0}^{2} - A^{2} x^{2}}$
Now $dU=\frac{2 m}{M}C_{V}dT$
By $1st$ law of thermodynamics
$dQ=dU+dW$
Here $dQ=0$ $\therefore dU=-dW$
or $\frac{2 m}{M}C_{v}dT=\frac{m}{M}RT\frac{2 A^{2} xdx}{V_{0}^{2} - A^{2} x^{2}}$
or $\frac{R}{\gamma - 1}dT=RT\frac{A^{2} xdx}{V_{0}^{2} - A^{2} x^{2}} \, $
or $\frac{1}{\gamma - 1}\displaystyle \int _{T_{0}}^{T}\frac{dT}{T}=A^{2}\displaystyle \int _{0}^{d}\frac{xdx}{V_{0}^{2} - A^{2} x^{2}}$
or $ \, \frac{1}{\gamma - 1}ln\frac{T}{T_{0}}=A^{2}\displaystyle \int _{0}^{d}\frac{xdx}{V_{0}^{2} - A^{2} x^{2}}$
Put $V_{0}^{2}-A^{2}x^{2}=z$ . Then $-2A^{2}xdx=dz$
$\therefore \displaystyle \int \frac{xdx}{V_{0}^{2} - A^{2} x^{2}}=\displaystyle \int -\frac{1}{2 A^{2}}\frac{dz}{z}=-\frac{1}{2 A^{2}}lnz=-\frac{1}{2 A^{2}}ln\left(V_{0}^{2} - A^{2} x^{2}\right)$
$\therefore \frac{1}{\gamma - 1}ln\frac{T}{T_{0}}=A^{2}\times -\frac{1}{2 A^{2}}\left(\left[ln \left(V_{0}^{2} - A^{2} x^{2}\right)\right]\right)_{0}^{d}$
$=-\frac{1}{2}\left[ln \left(V_{0}^{2} - A^{2} d^{2}\right) - lnV_{0}^{2}\right]$
$=-\frac{1}{2}\left[ln \left(1 - \frac{A^{2}}{V_{0}^{2}} \cdot \frac{\left(\right. n - 1 \left.\right)^{2} V_{0}^{2}}{\left(\right. n + 1 \left.\right)^{2} A^{2}}\right)\right]=-\frac{1}{2}ln\frac{4 n}{\left(n + 1\right)^{2}}$
or $ln\frac{T}{T_{0}}=-\frac{\gamma - 1}{2}ln\frac{4 n}{\left(n + 1\right)^{2}}$
or $T=T_{0}\left(\left[\frac{\left(\right. n + 1 \left.\right)^{2}}{4 n}\right]\right)^{\frac{\gamma - 1}{2}}$