REALLY NEED HELP 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? Use the correct number of significant digits. The initial temperature of the gas is

Explanation:

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

P= 1.83 atm

V = 2.50 L

n = 0.458 moles

Using an Ideal gas equation:

PV=nRT

T= PV/nR= 1.83*2.50/0.458* 0.0821 =121.81K

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

P' =2.05 atm

V = 2.50 L

n' = 0.458 mol + 0.713 mol = 1.171 mol

PV=nRT

T= PV/nR = 1.83*2.50/1.171*0.0802 = 53.37

The change in temperature will be

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.

The initial temperature of the gas is 122  K.

The final temperature of the gas is 53.3 K.

The temperature change is -68.7 K.

Explanation:

These are 100% Right answers on Edge

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