Heat - 2

Let us call final pressure (which will be the same fo both parts, since the system will be in equilibrium) as $P_f$ and the final volume of one of the parts as $V_f$ . Since both volumes add up initially to $6V$ , at the end they'll add also to $6V$ , so the final volume of the second part will be $6V - V_f$ . Thus, using Adiabatic process for an ideal gas equation:

$\large \displaystyle P(5V)^{1.5} = (P_f)(V_f)^{1.5}$ (i)

$\large \displaystyle (8P)V^{1.5} = (P_f)(6V-V_f)^{1.5}$ (ii)

Dividing (i) by (ii):

$\large \displaystyle \frac{5^{1.5}}{8} = \frac{V_f^ {1.5}}{(6V-V_f)^{1.5}}$

Raising both sides to $\frac23$ :

$\large \displaystyle \frac54 = \frac{V_f}{6V-V_f}$

$\color{#20A900} \boxed{ \large \displaystyle V_f = \frac{10}{3} V }$

Plugging this in (i):

$\large \displaystyle P(5V)^{1.5} = (P_f) \left( \frac{10}{3}V \right )^{1.5}$

$\color{#20A900} \boxed{ \large \displaystyle P_f = \frac32 \sqrt{\frac32} P }$

So:

$\color{#3D99F6} \boxed{ \large \displaystyle n = \frac32 \sqrt{\frac32} \approx 1.84 }$

As it is adiabatic process, here is no matter of concern about heat exchange between the two parts. Now, after being freed the pressure will adjust to get stability. Hence, on the both part pressure will be the same, let $P_0$ .

Now,

$P(5V)^{\gamma}=P_0{V_1}^\gamma$ .

And,

$(8P)V^{\gamma}=P_0{V_2}^\gamma$ .

From, these and substituting $\gamma=\frac{3}{2}$ we get,

$9P^{\frac{2}{3}}V=P^{\frac{2}{3}}(V_1+V_2)=P^{\frac{2}{3}}6V$ .

So, $P_0=(\frac{3}{2})^\frac{3}{2}P=1.837..P≈ 1.84P$ .