To determine the effect of a sudden increase in volume on the pressure of a container holding water, we can use the ideal gas law. However, please note that the ideal gas law is an approximation and may not be entirely accurate for liquids like water. Nonetheless, we can use it as an estimation.
The ideal gas law is given by:
PV = nRT
Where: P = Pressure V = Volume n = Number of moles of the substance R = Ideal gas constant T = Temperature
Since we are dealing with water, which is incompressible, we can assume that the number of moles remains constant. Therefore, we can rearrange the equation as follows:
P₁V₁ = P₂V₂
Where: P₁ = Initial pressure V₁ = Initial volume P₂ = Final pressure V₂ = Final volume
Let's calculate the final pressure after a volume increase of 0.05%:
P₁ = 200 bars (given) V₁ = Initial volume (unknown) P₂ = Final pressure (unknown) V₂ = V₁ + 0.05% of V₁
We want the final pressure (P₂) to be below 5 bars. Let's solve for the initial volume (V₁) that would yield a final pressure of 5 bars.
P₁V₁ = P₂(V₁ + 0.0005V₁) 200V₁ = 5(V₁ + 0.0005V₁) 200V₁ = 5V₁ + 0.0025V₁ 195V₁ = 0.0025V₁ 195V₁ - 0.0025V₁ = 0 194.9975V₁ = 0 V₁ = 0
From the equation, it seems that the initial volume required to reach a final pressure below 5 bars is zero. However, this result is mathematically incorrect and suggests that the assumptions made (such as the ideal gas law being applicable to liquids) may not be valid.
In reality, liquids like water are relatively incompressible, and a small increase in volume would not significantly affect the pressure. The ideal gas law may not provide an accurate prediction in this scenario.