A sample of Ar gas has a pressure of 1.20 atm at 125°C and a volume of 1.00 L. What would the temperature be (°C) with a pressure of 1.80 atm and constant volume?

597K

Explanation:

Given parameters:

Initial pressure = 1.2atm

Initial temperature = 125°C = 125 + 273 = 398K

Initial volume = 1.00L

Final pressure = 1.80atm

Final volume  = 1.00L

Unknown:

Final temperature = ?

Solution:

To solve this problem, we must understand that at constant volume, the pressure of a given mass of gas varies directly with the absolute temperature.

Mathematically;

              =

where P and T are the pressures

           1 and 2 are the initial and final states

Now insert the parameters and solve;

            =

         T₂ x 1.2 = 398 x 1.8

         T₂ = 597K

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

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