Air is contained in a cylinder fitted with a piston as below. The cross sectiona

Question

Air is contained in a cylinder fitted with a piston as below. The cross sectional area of the piston is 0.5 m2 and the gas is initially at a pressure of 2.0 atm and 300 K

A) A weight is placed on the piston that exerts a force equivalent to 10.0 atm resulting in the movement of the piston until it reaches a set of stops. Assume no heat transfer out of the container. What are the values of P, V and T at this point?

B)The air in the cylinder then slowly cools via heat transfer to the surroundings until it reaches a temperature of 300 K. What are the values of P, V and T at this point?

C)Calculate q, w, ?U and ?H for each of these processes.

D)Calculate q, w, ?U and ?H for the entire process.

Air is contained in a cylinder fitted with a piston as below. The cross sectional area of the piston is 0.5 m2 and the gas is initially at a pressure of 2.0 atm and 300 K A) A weight is placed on the piston that exerts a force equivalent to 10.0 atm resulting in the movement of the piston until it reaches a set of stops. Assume no heat transfer out of the container. What are the values of P, V and T at this point? B)The air in the cylinder then slowly cools via heat transfer to the surroundings until it reaches a temperature of 300 K. What are the values of P, V and T at this point? C)Calculate q, w, ?U and ?H for each of these processes. D)Calculate q, w, ?U and ?H for the entire process.

Explanation / Answer

The cross sectional area of the piston is 0.5 m2

gas is initially at a pressure of 2.0 atm and 300 K

A) weight force equivalent to 10.0 atm

P1 = 2 and T1 = 300K ....... V1 = 2*0.5 = 1m3

P2 = 10 and T2 = 300K......V2 = 1*0.5 = 0.5m3

T1*P2V2 / P1V1 = 300*10*0.5 / 2*1 = 750K

P, V and T at this point are 10atm, 0.5m3 and 750K respectively

B) Temperature only changing so P will be 10atm and T = 300K

V3 = T3*P2V2 /T2 P3 = T3*V2 /T2 = 300*0.5 / 750 = 0.2m3

P, V and T at this point are 10atm , 0.2m3 and 300K respectevily

C) w = P (delta V)+V (delta P) ....................w = nRT log(Vfinal /Vinitial) isothermal

.....deltaU (change in internal energy) = Q - W

....delta H = delta U + PdeltaV +VdeltaP

..................Process A - adiabatic

w = P (delta V)+V (delta P) =

P, V and T at this point are 2atm, 1m3 and 750K respectively

P2, V2 and T1 at this point are 10atm, 0.5m3 and 750K respectively

deltaP , delta V, are 8atm and 0.5 m3 respectively

w = 10*0.5+0.5*8 = 9

adiabatic deltaU = 0

q =delta U + w = 0 + 9 = 9

delta H = 9

.................Process B - ISOBARIC

w = P (delta V) = 10*(0.5-0.2) = 3

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