A heat-conducting piston can freely move inside a closed thermally insulated cylinder with an ideal gas. In equilibrium the piston divides the cylinder into two equal parts, the gas temperature being equal to `T_0`. The piston is slowly displaced. Find the gas temperature as a function of the ratio `eta` of the volumes of the greater and smaller sections. The adiabatic exponent of the gas is equal to `gamma`.

Explanation

Since here the piston is conducting and it is moved slowly the temperature on the two sides increases and maintained at the same value.
Elementary work done by the agent = Work done in compression - Work done in expansion
i.e., `dA = p_2 dV - p_1 dV = (p_2 - p_1) dV`
where `p_1` and `p_2` are pressures at any instant of the gas on expansion and compression side respectively.
From the gas law `p_1 (V_0 + Sx) = vRT` and `p_2(V_0 - Sx) = vRT`, for each section
(`x` is the displacement of the piston towards section `2` )
So, `p_2 - p_1 = vRT (2 Sx)/(V_0^2 - S^2 x^2) = vRT.(2V)/(V_0^2 - V^2)(as S_x = V)`
So `dA = vRT (2 V)/(V_0^2-V^2)dV `
Also, from the first law of thermodynamics
`dA = -dU = -2v( R)/(gamma - 1) dT (as dQ = 0)`
So, work done on the gas `= -dA = 2v. (R )/(gamma - 1) dT`
Thus `2v (R )/(gamma - 1) dT = vRT (2 V.dV)/(V_0^2 - V^2)`,
or, `(dT)/(T) = gamma - 1 (V dV)/(V_0^2 - V^2)`
When the left end is `eta` times the volume of the rigid end.
`(V_0 + V) = eta (V_0 - V)` or `V = (eta - 1)/(eta + 1) V_0`
On integrating `int_(T_0)^T (dT)/(T) = (gamma - 1) int_0^V (VdV)/(V_0^2 - V^2)`
or `1 n(T)/(T_0) =(gamma - 1)[-(1)/(2) 1n (V_0^2 - V^2)]_0^V`
=`-(gamma - 1)/(2) [1 n(V_0^2 - V^2)-1 n V_0^2 - V^2) - 1n V_0^2]`
=`(gamma - 1)/(2) [2n V_0^2 - 1n V_0^2{1 - ((eta - 1)/(eta + 1))^2}]=(gamma - 1)/(2) 1 n((eta + 1)^2)/(4 eta)`
Hence `T = T_0 (((eta + 1)^2)/(4 eta))^((gamma - 1)/(2))`.
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