Respuesta :
Answer:
The answers to the question are
(A)
- (1) Q = 1430 J , W = - 405.3 J
- (2) Q = 1357.8 J, W = 0 J
- (3) Q = - 273.89 J, W 273.89 J
- (4) Q =339.614 J , W = 0
(B) Work done = -131.61 J
(C) Cycle efficiency is given by = 9.2 % less efficient than Carnot cycle that has 44.44 % efficiency
Explanation:
(1) For constant pressure
Q = m×cp×(T₂-T₁) and
W = -p×(v₂-v₁)
T₁/v₁ =T₂/v₂ or v₂ = T₂v₁/T₁
v₂ = 6 L
Work W = - 2 atm×(6 L-4 L) = -202650 Pa×0.002 m³ = - 405.3 J
R = cp×(Y-1)/Y = 285.7 J/(kg K)
and Q = m×1.005×(450-300)
pv = RT
pv/(RT) = 202650×0.004/(285.7×300) = = 9.46×10^(-3) kg
Q = 9.46×10⁻³×1.005×150 = 1430 J
(2) For constant volume we have
p₁/T₁ =p₂/T₂
Therefore p₂ = p₁T₂/T₁ where T₁ = 450 K and T₂ = 250 K
p₂ = 202650×250/450 = 112583.33 Pa
Q = m×cv×(T₂-T₁) and W = 0
Therefore cv = 1.005/1.4 = 0.718 and Q = 9.46×10³×0.718×200 = 1.3578 kJ
= 1357.8 J
(3) For constant temperature we have
W = p₁v₁×[tex]ln(\frac{v_{1} }{v_{2} } )[/tex] per unit mass of the gas
Q = - W
p₁v₁ =p₂v₂ or p₂ = p₁v₁/v₂ where p₁ = 112583.33 Pa
p₂ = 112583.33 Pa ×1.5 = 168874.995 Pa
v₁ = 6 L, v₂ = 4 L
and W = 112583.33 Pa × 0.006 m³ × [tex]ln(\frac{v_{1} }{v_{2} } )[/tex] = 675500 × ㏑(3/2) = 273.89 J
Q = - 273.89 J
(4) For constant volume process we have
For constant volume we have
p₁/T₁ =p₂/T₂ where p₁ = 168874.995 Pa
Therefore p₂ = p₁T₂/T₁ where T₁ = 250 K and T₂ = 300 K
p₂ = 168874.995×300/250 = 202649.994 Pa
Q = m×cv×(T₂-T₁) and W = 0
Therefore cv = 1.005/1.4 = 0.718 and Q = 9.46×10⁻³×0.718×50 = 339.614 J
(B) net work = -405.3 J +273.89 J = -131.61 J
(C) Cycle efficiency is given by
η = -W/Q₁ = (131.61 J/1430 J ) × 100 = 9.2 %
For Carnot cycle we have η (Carnot) = 1 - T₂/T₁ = 1-250/450 = 0.4444
or 44.44 %
The cycle is less efficient
