Question

A container holds 30m of air at a pressure of 150000Pa. If the volume changed to 10m by decreasing load on the piston. What will the pressure of the gas become? Assume that its temperature remains constant.​

Answer 1

Answer:

100000

Explanation:

150000/30

=50000

150000/10

=100000

Answer 2

Answer:

The pressure of the gas after the load on the piston decreased is equal to 45×10⁴ Pa.

Explanation:

Given, the initial volume of the air, V₁ = 30m³

The initial pressure of the air, P₁ = 15 × 10⁴Pa

When the load decreased, the volume of the air will be:

The final volume of the air, V₂ = 10m³

Consider that P₂ is the final pressure of air.

According to Boyle's law:

The product of the pressure and volume of the given mass of gas is constant at constant temperature.

PV = constant

P₁V₁ = P₂V₂

$P_2=\frac{P_1V_1}{V_2}$

$P_2=\frac{(30)(15*10^4)}{10}$

$P_2=45\times 10^4Pa$

Therefore, the final pressure of the air is equal to 45×10⁴ Pa.

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