A sample of argon gas occupies 100 L at 720. mmHg and 20°C. What will
be the temperature of the argon gas at 360. mm Hg and 2.14 L. assuming
the number of moles of gas remained constant?

T₂ =  3.135 K

Explanation:

Given data:

Initial volume = 100 L

Initial pressure = 720 mmHg

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = ?

Final volume = 2.14 L

Final pressure = 360 mmHg

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

P₁V₁/T₁ = P₂V₂/T₂  

T₂   = P₂V₂T₁ /P₁V₁

T₂ = 360 mmHg × 2.14 L × 293 K / 720 mmHg × 100 L  

T₂ = 225727.2 mmHg .L. K / 720 mmHg × 100 L  

T₂ =  3.135 K

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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