A gas has a temperature of 273.15 k and a pressure of 101.325 kpa. what can be concluded about the gas?

We are given with a temperature of 273.15 Kelvin and a pressure of 101.325 kPa. Using PV=nRT rule of ideal gas, where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature, we can conclude that the conditions are under STP(standard temp and pressure) and one mole of this gas occupies 22.4 L of volume.

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Alleei Alleei · Answer 2 · 2019-10-31T09:24:36+01:00

Answer : A gas has a temperature of 273.15 K and a pressure of 101.325 kPa from this we conclude that the gas reached at standard temperature and pressure.

Explanation :

STP : STP stands for standard temperature and pressure.

At STP,

The temperature is, [tex]0^oC[/tex] or [tex]273.15K[/tex] or [tex]32^oF[/tex]

The pressure is, [tex]1atm[/tex] or [tex]101.325kPa[/tex]

The volume is, 22.4 L

As per question we conclude that, when the temperature and pressure of the gas will be 273.15 K and 101.325 kPa respectively this means that the gas reached at standard temperature and pressure.