A gas storage tank has a volume of 3.5 x 10^5 m^3 when the temperature is 27 C and the pressure is 101 kPa. What os the new volume of the tank if the temperature drops to -10 C and the pressure drops to 95 kPa

3.26×10⁵m³

Explanation:

Given data:

Initial volume = 3.5×10⁵ m³

Initial temperature = 27 °C (27+273= 300 K)

Initial pressure = 101 Kpa

Final volume = ?

Final temperature = -10°C (-10+273 = 263 K)

Final pressure = 95 Kpa

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 101 Kpa × 3.5×10⁵ m³× 263 K / 300 K × 95 Kpa  

V₂ = 92970.5 ×10⁵kpa. m³. K / 28500 K.Kpa

V₂ = 3.26×10⁵m³

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.