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?

he 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:

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

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.

Explanation:

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

the answer is c. nacl