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 final temperature of the gas is K.

Answer 1

Answer:The change in temperature of the gas mixture is -68.44 K.

Explanation:

1. Temperature of the bottle when only helium gas was present :T

Pressure inside the container =P= 1.83 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n = 0.458 moles

Using an Ideal gas equation:

$PV=nRT$

$T=\frac{PV}{nR}=\frac{1.83 atm\times 2.50 L}{0.458 mol\times 0.0820 atm L/mol K}=121.81 K$

2. Temperature of the bottle when argon gas is added into the container:T'

Pressure inside the container = P' =2.05 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n' = 0.458 mol + 0.713 mol = 1.171 mol

$P'V=n'RT'$

$T'=\frac{P'V}{n'R}=\frac{2.05 atm\times 2.50 L}{1.171 mol\times 0.0820 atm L/mol K}=53.37 K$

The change in temperature will be given as:

Final temperature = Initial temperature:T' - T

53.37 k - 121.81 K = -68.44 K

The change in temperature of the gas mixture is -68.44 K.

Answer 2

Answer:

Initial Temp is 122 K

Final Temp 53.3 K

Temp Change -68.7