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

Answer: 4

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
$=F_{agent}dx=\left(p_2-p_1\right)Adx$
or $dW=\frac{m}{M}RT\left(\frac{1}{V_0-Ax}-\frac{1}{V_0+Ax}\right)Adx$
$=\frac{m}{M}RT\frac{2A^{2}xdx}{V_0^2-A^2{x}^2}$
Now $dU=\frac{2m}{M}{C}_{V}dT$
By $1st$ law of thermodynamics
$dQ=dU+dW$
Here $dQ=0$ $\therefore dU=-dW$
or $\frac{2m}{M}{C}_{v}dT=\frac{m}{M}RT\frac{2A^{2}xdx}{V_0^2-A^2{x}^2}$
or $\frac{R}{\gamma -1}dT=RT\frac{A^{2}xdx}{V_0^2-A^2{x}^2}$
or $\frac{1}{\gamma -1}{\int }_{T_0}^T\frac{dT}{T}=A^2{\int }_0^d\frac{xdx}{V_0^2-A^2{x}^2}$
or $\frac{1}{\gamma -1}ln\frac{T}{T_0}=A^2{\int }_0^d\frac{xdx}{V_0^2-A^2{x}^2}$
Put $V_0^2-A^2{x}^2=z$ . Then $-2A^{2}xdx=dz$
$\therefore \int \frac{xdx}{V_0^2-A^2{x}^2}=\int -\frac{1}{2A^2}\frac{dz}{z}=-\frac{1}{2A^2}lnz=-\frac{1}{2A^2}ln\left(V_0^2-A^2{x}^2\right)$
$\therefore \frac{1}{\gamma -1}ln\frac{T}{T_0}=A^2\times -\frac{1}{2A^2}{\left(\left[ln\left(V_0^2-A^2{x}^2\right)\right]\right)}_0^d$
$=-\frac{1}{2}\left[ln\left(V_0^2-A^2{d}^2\right)-ln{V}_0^2\right]$
$=-\frac{1}{2}\left[ln\left(1-\frac{A^2}{V_0^2}\cdot \frac{(n-1{)}^2{V}_0^2}{(n+1{)}^2{A}^2}\right)\right]=-\frac{1}{2}ln\frac{4n}{{\left(n+1\right)}^2}$
or $ln\frac{T}{T_0}=-\frac{\gamma -1}{2}ln\frac{4n}{{\left(n+1\right)}^2}$
or $T=T_0{\left(\left[\frac{(n+1{)}^2}{4n}\right]\right)}^{\frac{\gamma -1}{2}}$