Question

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture? The initial temperature of the gas is K.

Answer 1

122 K

Step-by-step explanation:

Answer 2

Answer:The change in temperature of the gas mixture is -68.43 K.

Initial temperature of the gas 121.81 K.

Step-by-step explanation:

Initial temperature of the container with 2.50 L of volume of helium gas alone:T

Initial pressure ,P = 1.83 atm

number of moles of helium ,n= 0.458 mol

V = 2.50 L

`PV=nRT` ( Ideal gas equation)

R = universal gas constant = 0.0820 atm L/mol K

`T=121.81 K`

Initial temperature of the gas 121.81 K.

Final temperature of the container with 2.50 L of volume with helium and argon gas:T'

Final pressure ,P' = 2.05 atm

Number of moles of helium ,n'= 0.713 mol + 0.458 mol = 1.171 mol[/tex]

V = 2.50 L

`P'V=n'RT'`

`T'=53.37 K`

The change in temperature of the gas mixture
`\Delta T =T'-T=53.37 K-121.81 k =-68.43 K`