Asample of oxygen occupies a volume of 160dm3 at 91 degrees celsius. what will be the volume of the oxygen when the temperature drops to 0.00 degrees celsius

V₂ = 120 dm³

Explanation:

Given data:

Volume of oxygen = 160 dm³

Initial temperature = 91 °C

Final volume = ?

Final temperature = 0 °C

Solution:

First of all we will convert the temperature into kelvin.

91 + 273 = 364 k

0 + 273 = 273 K

According to Charle's law,

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 160 dm³ × 273 k / 364 k

V₂ = 43680 dm³.K / 364 k

V₂ = 120 dm³

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a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.