Respuesta :
Answer:
Q = E + W = 0.25 + 0.4 = 0.29KJ
W = 0.4KJ
b) 2000KJ
Explanation:
The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.
mathematically from first law,
dE = dQ - dW
where E is internal energy
Where energy may be transfereed as eitherf work (W) or heat (Q)
Work on the other hand can be at constant volume and pressure.
at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit
Work = p(Vf -Vi)
attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ
from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ
To get the heat transfer, from dE = dQ - dW
Q = E + W = 0.25 + 0.4 = 0.29KJ
b) for the second question ; from Pdv = p(Vf -Vi), but the pressure here is in atmosphere
W = 100000 ( 0.12 - 0.1) = 2000KJ
Answer: (a). W = 4KJ and Q = 4.25KJ
(b). W = -2KJ and ΔPE = 2KJ
Explanation:
(a).
i. We are asked to calculate the work done during the expansion process considering gas as system.
from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁
so W = P(V₂-V₁)
W = (2 × 10²) (0.12 - 0.10)
W = 4 KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0
∴ Q - W = ΔU
Q = ΔU + W
Q = 0.25 + 4
Q = 4.25 KJ
(b).
i. to calculate the work done during the expansion process considering piston as system;
W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.
W = (Patm - Pgas)(V₂ - V₁)
W = (1-2) ×10² × (0.12-0.1)
W = -2KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
Q = 0 since the piston and cylinder walls are perfectly insulated.
for piston, we neglect the change in internal energy and kinetic energy
ΔU = ΔKE = 0
from Q - W = ΔU + ΔKE + ΔPE
0 - (-2) = 0 + 0 + ΔPE
∴ ΔPE = 2KJ
