Chemistry, 21.11.2019 08:31 kaileyy06

Arigid, 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.

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Answer from: larueeee25

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 from: foralways3178

The temperature of the bottle has been dropped by 68.29 K, and the final temperature of the bottle has been 53.385 K.

The gas has been assumed to be an ideal gas. Thus according to the ideal gas equation:

PV = nRT

P = Pressure of He gas = 1.83 atm

V = Volume of He gas = 2.50 L

n = moles of gas = 0.458

R = constant = 0.0821

T = temperature

The temperature of the bottle in the presence of He gas has been:

1.83 $\times$ 2.50 = 0.458 $\times$ 0.0821 $\times$ Temperature

4.575 = 0.0376 Temperature

Temperature = 121.675 K.

On the addition of Argon (0.713 mol)

n = 0.713 + 0.458 mol

n = 1.171 mol

The pressure of the bottle has been increased to 2.05 atm.

P = 2.05 atm

Substituting the values, the temperature of the bottle after the addition of Argon:

2.05 $\times$ 2.50 = 1.171 $\times$ 0.0821 $\times$ Temperature

5.125 = 0.096 Temperature

Temperature = 53.385 K.

The change in temperature can be given by:

Change in temperature = Final temperature - Initial temperature

Change in temperature = 53.385 K - 121.675 K

Change in temperature = -68.29 K

The temperature of the bottle has been dropped by 68.29 K, and the final temperature of the bottle has been 53.385 K.

For more information about the change in temperature, refer to the link:

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Arigid, 2.50 l bottle contains 0.458 mol he. the pressure of the gas inside the bottle is 1.83 atm....

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