Question

If we have 5m3 of helium at 500K and 50,000Pa, what are the final pressure and temperature, an the change of energy, heat added, and work done if the volume doubles a) isobarically, b) Isothermally, and c) adiabatically?

Answer 1

given

P1 = 50000 Pa

V1 = 5 m^3

T1 = 500 K

from ideal gas equation

P1*V1 = n*R*T1

number moles n = P1*V1/(R*T1)

n = 50000*5/(8.314*500)

n = 60.14 moles

(a)

isobaric

In isobaric pressure remains same

P2 = P1 = 50000 Pa

V2 = 2*V1 = 10 m^3

T2/T1 = V2/V1

T2/500 = 10/5

T2 = 1000 K

change in energy dU = n*CV*dT = n*(3/2)*R*(T2-T1)

change in energy dU = 60.14*(3/2)*8.314*(1000-500) = 375000 J

heat added = Q = n*Cp*dT = n*(5/2)*R*dT = 625000 J

work done = Q - dU = 250000 J

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isobaric

In isobaric temperature remains same

T2 = T1 = 500 K

V2 = 2*V1 = 10 m^3

P2/P1 = V1/V2

P2/50000 = 5/10

P2 = 25000 Pa

change in energy dU = n*CV*dT = n*(3/2)*R*(T2-T1)

change in energy dU =0

heat added = Q = work = 2.303*n*R*T1*log(V2/V1)

heat added = Q = 2.303*60.14*8.314*500*log(10/5)

heat added = Q = 173319.4 J

work done = 173319.4 J

part(c)

In adiabatic process

total heat remains same

V2 = 10 m^3

from adiabatic relation

P1*V1^gamma = P2*V2^gamma

50000*5^(5/3) = P2*10^(5/3)

P2 = 15749 Pa

T1*V1^(gamma-1) = T2*V2^(gamma-1)

500*5^(5/3-1) = T2*10^(5/3-1)

T2 = 315 K

change in energy dU = n*CV*dT = n*(3/2)*R*(T2-T1)

change in energy dU = 60.14*(3/2)*8.314*(315-500) = -138751.1 J

heat added = Q = 0

work done = -dU = 138751.1 J