1.20 L of oxygen gas is held in a flexible vessel at STP. If the
vessel is transported to the bottom of Pacific Ocean where the
pressure is 100 atm and the temperature is 15 °C, what is the
new volume of the gas?

The new volume of the gas when the vessel is transported to the bottom of the Pacific Ocean is 0.013 L

From the question given above, the following data were obtained:

Initial volume (V₁) = 1.2 L

Initial pressure (P₁) = STP = 1 atm

Initial temperature (T₁) = STP = 273 K

New pressure (P₂) = 100 atm

New temperature (T₂) = 15 °C = 15 + 273 = 288 K

Using the combined gas equation, the new volume of gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

(1 × 1.2) / 273 = (100 × V₂) / 288

1.2 / 273 = (100 × V₂) / 288

Cross multiply

273 × 100 × V₂ = 1.2 × 288

27300 × V₂ = 345.6

Divide both side by 27300

V₂ = 345.6 / 27300

Therefore, the new volume of the gas is 0.013 L

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New Volume = 0.013 Liter

Explanation:

Combined Gas Law Problem

P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1 atm

V₁ = 1.2 Liters

T₁ = 0°C = 273 K

P₂ = 100 atm

V₂ = V₁(P₁/P₂)(T₂/T₁)

T₂ = 15°C = (273 + 15)K =   288K

P₂ = 1.2L(1atm/100atm)(288K/273K) = 0.013 Liter

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the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.