A cylinder contains oxygen gas at a pressure of 2.00 atm. The volume is 4.00 L, and the temperature is 300 K. Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes:

(1) Heated at constant pressure from the initial state (state 1) to state 2, which has 450 K
(2) Cooled at constant volume to 250 K (state 3).
(3) Compressed at constant temperature to a volume of 4.00 L (state 4).
(4) Heated at constant volume to 300 K, which takes the system back to state 1.

A) Calculate Q and W for each of the four processes.

B) Calculate the net work done by the oxygen.

C) What is the effiency of this device as a heat engine? How does this effiiency compare with that of a Carnot-cycle engine operating between the same minimum temperatures of 250K and 450K?

The answers to the  question are

(A)  

(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× 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 ×  = 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

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

answer: the correct answer for the fill in the blank is rods.

rods are the specialized photoreceptor cells that are present in the retina of human eye. they are responsible for vision in the night time ( also called scotopic vision). they have very low visual acuity and color distinguishing ability.

as they are responsible for vision in night time, therefore they are found predominantly in the nocturnal animals.