Question

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Answer 1

The new pressure will be the same when the piston is left. Let this pressure be P'.

The total volume of the gas is 6V. After leaving the piston, let the new volume of first part be V' so that the volume of other part will be (6V - V').

For adiabatic process, we know that,

$\longrightarrow\sf{P_i\left(V_i\right)^{\gamma}=P_f\left(V_f\right)^{\gamma}}$

So for the first part,

$\longrightarrow\sf{P(5V)^{\gamma}=P'(V')^{\gamma}\quad\quad\dots(1)}$

And for the second part,

$\longrightarrow\sf{8PV^{\gamma}=P'(6V-V')^{\gamma}\quad\quad\dots(2)}$

Dividing (2) by (1),

$\longrightarrow\sf{\dfrac{8PV^{\gamma}}{P(5V)^{\gamma}}=\dfrac{P'(6V-V')^{\gamma}}{P'(V')^{\gamma}}}$

$\longrightarrow\sf{\left(\dfrac{6V-V'}{V'}\right)^{\gamma}=\dfrac{8}{5^{\gamma}}}$

Here, $\sf{\gamma=1.5=\dfrac{3}{2}.}$

$\longrightarrow\sf{\left(\dfrac{6V-V'}{V'}\right)^{\frac{3}{2}}=\dfrac{8}{5^{\frac{3}{2}}}}$

$\longrightarrow\sf{\dfrac{6V-V'}{V'}=\dfrac{8^{\frac{2}{3}}}{5}}$

$\longrightarrow\sf{\dfrac{6V-V'}{V'}=\dfrac{4}{5}}$

$\longrightarrow\sf{\dfrac{6V-V'}{V'}=\dfrac{9-5}{5}}$

By rule of dividendo,

$\longrightarrow\sf{\dfrac{V'}{6V}=\dfrac{5}{9}}$

$\longrightarrow\underline{\underline{\sf{V'=\dfrac{10}{3}\,V}}}$

And so,

$\longrightarrow\underline{\underline{\sf{6V-V'=\dfrac{8}{3}\,V}}}$

From (1),

$\longrightarrow\sf{P(5V)^{\frac{3}{2}}=P'\left(\dfrac{10}{3}\,V\right)^{\frac{3}{2}}}$

$\longrightarrow\underline{\underline{\sf{P'=P\left(\dfrac{3}{2}\right)^{\frac{3}{2}}}}}$

For isothermal process, we know that,

$\longrightarrow\sf{P_iV_i=P_fV_f}$

So for the first part,

$\longrightarrow\sf{5PV=P'V'\quad\quad\dots(3)}$

And for the second part,

$\longrightarrow\sf{8PV=P'(6V-V')\quad\quad\dots(4)}$

Dividing (4) by (3),

$\longrightarrow\sf{\dfrac{8PV}{5PV}=\dfrac{P'(6V-V')}{P'V'}}$

$\longrightarrow\sf{\dfrac{6V-V'}{V'}=\dfrac{8}{5}}$

$\longrightarrow\sf{\dfrac{6V-V'}{V'}=\dfrac{13-5}{5}}$

By rule of dividendo,

$\longrightarrow\sf{\dfrac{V'}{6V}=\dfrac{5}{13}}$

$\longrightarrow\underline{\underline{\sf{V'=\dfrac{30}{13}\,V}}}$

And so,

$\longrightarrow\underline{\underline{\sf{6V-V'=\dfrac{48}{13}\,V}}}$

From (3),

$\longrightarrow\sf{5PV=\dfrac{30}{13}\,P'V}$

$\longrightarrow\underline{\underline{\sf{P'=\dfrac{13}{6}P}}}$