⇒ A gas ‘G’ is contained in a vessel of capacity 40 litres and subjected to a pressure of 20 atm. When this container is connected to a second empty container of the same capacity, find the pressure that will be acting on the gas ‘G’ in both the containers if the temperature is kept constant.. Created by CopyThat. Chemistry

A gas ‘G’ is contained in a vessel of capacity 40 litres and subjected to a pressure of 20 atm. When this container is connected to a second empty container of the same capacity.The pressure that will be acting on the gas ‘G’ in both the containers if the temperature is kept constant.Initial conditions:Final conditions:Since Temperature is kept constant,Applying Boyle's law⭐________________________⭐

⇒ A gas ‘G’ is contained in a vessel of capacity 40 litres and subjected to a p...

XxItsDivYanShuxX

$\LARGE\bold { \color{magenta}{ \underline{ \color{yellow}{\mathcal{Given:↓}}}}}$

A gas ‘G’ is contained in a vessel of capacity 40 litres and subjected to a pressure of 20 atm.
When this container is connected to a second empty container of the same capacity.

$\LARGE\bold { \color{magenta}{ \underline{ \color{lime}{\mathcal{Find:↓}}}}}$

The pressure that will be acting on the gas ‘G’ in both the containers if the temperature is kept constant.

$\LARGE\bold { \color{magenta}{ \underline{ \color{aqua}{\mathcal{Solution:↓}}}}}$

Initial conditions:

$\bold{ V_1= 40 \: litres}$

$\bold{P_1 = 20 \: atm}$

Final conditions:

$\bold{V_2 = (40 + 40) = 80 \: litres}$

$\bold{P_2 = \: ? }$

Since Temperature is kept constant,

Applying Boyle's law

$\bold{P_1V_1 = P_2V_2}$

$\bold{20 \times 40 = P_2 \times 80}$

$\bold{P_2 = \frac{20 \times 40}{80} \implies \frac{800}{80} } \implies \Large\bold{ \color{magenta}{ \boxed{10 \: atm }}}$

$\bold{ \therefore \: The \: pressure \: of \: the \: gas \: in \: both \: the \: containers}\implies \Large\bold{ \color{navy}{ \boxed{10 \: atmosphere }}}$

⭐________________________⭐

XxMrElash25xX

Capacity of vessel = 9 L

Capacity of barrel = 1 L

When the piston is lifted for first time, the air flows from vessel to the barrel and volume occupied by air will be total 10 L.

According to Boyle's law new pressure will be:

P1V1 = P2V2

76×9 = P2×10

New pressure, P2 = 68.4 cm of Hg

Now when the piston is lowered, the air from barrel is released. But pressure inside the vessel will remain same. When piston makes the 2nd upward stroke, the air from vessel travels to the barrel and volume occupied by air will be again 10L. Now new pressure will be:

P1V1 = P2V2

68.4×9 = P2×10

New pressure after the piston of the barrel makes 2 upward strokes, P2 = 61.56 cm of Hg

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Initial conditions:

Final conditions:

Since Temperature is kept constant,

Applying Boyle's law

⭐________________________⭐

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