Two cylinders each contain 0.20 mol of a diatomic gas at 350 K and a pressure of

Question

Two cylinders each contain 0.20 mol of a diatomic gas at 350 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is 1.0 atm.

Part A

What is the final temperature of the gas in the cylinder A?

Express your answer to two significant figures and include the appropriate units.

Part B

What are the final temperature of the gas in the cylinder B?

Express your answer to two significant figures and include the appropriate units.

Part C

What is the final volume of the gas in the cylinder A?

Express your answer to two significant figures and include the appropriate units.

Part D

What is the final volume of the gas in the cylinder B?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Cylinder A: First one is Isothermal. T does not change. T=350K
V=(nRT)/P = (.2mol x 8.31 x 350K)/101325Pa = 5.74*10^-3 m^3

Cylinder B: Adabiatic, a little more complicated.
You can find V by using PV^gamma = constant
In this case its a diatomic gas which means gamma is 1.4.
The reason this works is because PV^gamma(before process) =PV^gamma(after process)
Using V=nRT/P for before process = (.2mol x 8.31 x 350K)/303975Pa = 1.9136*10^-3
PV^gamma=PV^gamma=> 303975Pa x (1.9136*10^-3m^3)^1.4=303975Pa x 1.565*10^-4
Solving for V you get 4.758*10^-3m^3
And Then PV/nR=T
T=290.07K

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