# The gas is compressed isometrically from the volume V1 = 8 liters to the volume V2 = 6 liters.

**The gas is compressed isometrically from the volume V1 = 8 liters to the volume V2 = 6 liters. At the same time, the pressure increased by p = 4 kPa. What was the initial pressure p?**

The gas initially had a pressure p₁ and a volume V₁ = 8 l = 0.008 m ^ 3.

The gas is compressed isothermally (T = const) to a volume of V₂ = 6 l = 0.006 m ^ 3, the pressure increased by Δ p = 4 kPa = 4000 Pa to a certain value p₂ = p₁ + Δ p.

To determine what the initial pressure p₁ was, we use the Boyle-Mariotte law:

р₁ ∙ V₁ = р₂ ∙ V₂ or р₁ ∙ V₁ = (р₁ + Δ р) ∙ V₂, we get,

р₁ ∙ (V₁ – V₂) = Δ р ∙ V₂ or р₁ = Δ р ∙ V₂ / (V₁ – V₂).

Substitute the values of the quantities into the calculation formula:

p₁ = 4000 Pa ∙ 0.006 m ^ 3 / (0.008 m ^ 3 – 0.006 m ^ 3).

p₁ = 12000 Pa = 12 kPa

Answer: the initial gas pressure was 12 kPa