The pressure of 6.0 L of a gas is 200 kPA. What will be the volume of the pressure is doubled, keeping the temperature constant?

3. 0 L

Explanation:

We'll use P₁V₁/T₁ = P₂V₂/T₂

Because T₁ and T₂ are constant throughout, then the two cancel out and we are left with;

P₁V₁ = P₂V₂

We can bow replace the variables with given values form the question;

200 * 6.0 = 400 * V₂

200 * 6 / 400 = V₂

3 = V₂

= 3.0 L