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If a 2 liter container with an initial pressure of 2 atms is crushed to half its original volume what would its new pressure be in kPa’s

Respuesta :

The final pressure will be 405.3 kilopascals.

Explanation:

According to the Boyle's Law, the pressure of the gas is inversely proportional to the volume of the gas as the temperature of the gas remains constant.

So,PV = Constant.

Here in this question, the volume of the container is halved.

Let the Pressure of the gas initially be P1 and final pressure be P2. Volume of gas initially be V

So, according to Boyle's Law,

P₁V₁ = P₂V₂PP.

So, [tex]2 \times V[/tex] = P₂[tex]\frac V2[/tex].

So, P2 = 4 atm.

1atm = 101.325 kilo pascals.

So, 4 atm = [tex]4\times101.325[/tex] kilopascals. = 405.3 kilopascals.

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