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A bottle at 317 K contains an ideal gas at a pressure of 150 kPa. The...

A bottle at 317 K contains an ideal gas at a pressure of 150 kPa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against Pext = 107 kPa, and some gas is expelled from the bottle in the process. When P = Pext, the stopper is quickly replaced. The gas remaining in the bottle slowly warms to 317 K. What is the final pressure (in kPa) in the bottle for a monatomic gas for which CV,m = 3R/2, and a diatomic gas for which CV,m = 5R/2?

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Answer #1

Part a

For monoatomic gas

The given process is adiabatic

Q = 0

From the first law of thermodynamics

Q = U - W = 0

U = W

Change in internal energy = work done

nCv(T2-T1) = - Pext(V2 - V1)

From the ideal gas equation

nCv(T2-T1) = - Pext(nRT2/P2 - nRT1/P1)

Cv(T2-T1) = - RPext(T2/P2 - T1/P1)

(3R/2)(T2 - 317) = - R*107*10^3Pa(T2/107*10^3Pa - 317/150*10^3Pa)

1.5(T2 - 317) = -(T2 - 226.12)

1.5T2 - 475.5 = - T2 + 226.12

2.5 T2 = 701.62

T2 = 280.65 K

After stopper is replaced

T2 = 280.65 K

P2 = Pext = 107*10^3Pa

T3 = 317 K

P3 =?

P3/T3 = P2/T2

P3 = 107*10^3Pa * 317K / 280.65K

P3 = Pf = 120858 Pa = 121 kPa

Part b

For diatomic gas

The given process is adiabatic

Q = 0

From the first law of thermodynamics

Q = U - W = 0

U = W

Change in internal energy = work done

nCv(T2-T1) = - Pext(V2 - V1)

From the ideal gas equation

nCv(T2-T1) = - Pext(nRT2/P2 - nRT1/P1)

Cv(T2-T1) = - RPext(T2/P2 - T1/P1)

(5R/2)(T2 - 317) = - R*107*10^3Pa(T2/107*10^3Pa - 317/150*10^3Pa)

2.5(T2 - 317) = -(T2 - 226.12)

2.5T2 - 792.5 = - T2 + 226.12

3.5 T2 = 1018.62

T2 = 291.03 K

After stopper is replaced

T2 = 291.03 K

P2 = Pext = 107*10^3Pa

T3 = 317 K

P3 =?

P3/T3 = P2/T2

P3 = 107*10^3Pa * 317K / 291.03K

P3 = Pf = 116548 Pa = 117 kPa

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