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

given

n = 0.2 mol
T1 = 350 k
P1 = 3 atm = 3*1.013*10^5 = 3.065*10^5 Pa

P2 = 1 atm = 1.013*10^5 pa

Use P1*V1 = n*R*T1

==> V1 = n*R*T1/P1

= 0.2*8.314*350/(3.065*10^5)

= 0.0019 m^3

A) In Isothermal process temperature is constant.

so, Final temperature is 350 K

B)
P^(1-gamma)*T^gamma = constant

P2^(1-gamma)*T2^gamma = P1^(1-gamma)*T1^gamma

T2^gamma = (P1/P2)^(1-gamma)*T1^gamma

T2 = T1*(P1/P2)^(1/gamma - 1)

= 350*(3/1)^(1/1.4 - 1)

= 256 k

C) In Isothermal process, P2*V2 = P1*V1

==> V2 = P1*V1/P2

= 3*0.0019/1

= 0.0057 m^3

D) In adiabatic process, P2*V2^gamma = P1*V1^gamma

V2 = V1*(P1/P2)^(1/gamma)

= 0.0019*(3/1)^(1/1.4)

= 0.0042 m^3

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