Helium Gas Has A Volume Of 250 At 273k At 1 Atm.What Will Be The Final Pressure If The Volume Is Reduced To 100l At 318 K

Helium gas has a volume of 250 at 273k at 1 atm.what will be the final pressure if the volume is reduced to 100L at 318 k

 

Given Data:

V1 = 250 L                 V2 = 100L

T1 = 273K                   T2 = 318K

P1 = 1 atm                   P2 = unknown ?

Solution:

In this problem we can used the formula of combined gas law which is

V1P1/T1 = V2P2/T2

To simplify we can used the formula

P2 = V1×P1×T2 / T1×V2

P2 = 250L × 1 atm × 318K / 273K × 100L

P2 =  79, 500 atm / 27, 300

P2 = 2.912 atm

The final pressure is 2.912 atm

note: make sure to cancel the unit of Kelvin and Liter

Hope this helps...


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