Question

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

Answer 1

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, $2 \times V$ = P₂$\frac V2$.

So, P2 = 4 atm.

1atm = 101.325 kilo pascals.

So, 4 atm = $4\times101.325$ kilopascals. = 405.3 kilopascals.