A cylinder contains oxygen gas (O2) at a pressure of 2.40 atm. The volume is 4.0

Question

A cylinder contains oxygen gas (O2) at a pressure of 2.40 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 440 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.

Part A
Calculate Q for each of the four processes.
Answer in the order indicated. Separate your answers using commas .
Q_1 to 2,  Q_2 to 3,  Q_3 to 4,  Q_4 to 1 = ___ J

Part B
Calculate W for each of the four processes.
W_1 to 2,  W_2 to 3,  W_3 to 4,  W_4 to 1 = ___ J

Part C
Calculate the net work done by the oxygen.
W = ___ J

Part D
What is the efficiency of this device as a heat engine?
e = ___ %

Explanation / Answer

Given that, Initial pressure P1 = 2.40 atm Initial volume V1 = 4.00 L Initial temperature T1 = 300 K 1) Temperature T2 = 440 K      Ideal gas equation PV = nRT    At constant pressure (V1 / T1)  = (V2 / T2)                                       (4 / 300) = (V2 / 440)                                                V2 = 5.86 L 2) Temperature T3 = 250 K    At constant volume (P1 / T2 ) = (P2 / T3 )                                           P2 = 1.36 atm 3) Volume V3 = 4.00 L At constant temperature P2V2 = P3V3                                          P3 = 1.99 atm 4) Temperature T4 = 300 K At constant volume (P3 / T3 ) = (P4 / T4 )                                        P4 = 2.38 atm PART A: For an Ideal gas Q is calculated by Q = U + W     U = (3/2) nRT and W = PV Here, R = 8.315 J/mol.K          n = 1    Q1 to 2 = U1 to 2 + W1 to 2                              = 1746.15 J + 4.46 J                = 1750.61 J    Q2 to 3 = U2 to 3 + W2 to 3                           = -2369.775 + 0 J               = -2369.775 J Q3 to 4 = U3 to 4 + W3 to 4              = 632.625 + (-3.7014) J              = 628.923 J Q4 to 1 = U4 to 1 + W4 to 1              = 0 J + 0 J              = 0 J PART B:                 = 0 J + 0 J              = 0 J PART B:    W1 to 2 = P1 V = P1 ( V2 - V1 )             = 4.46 J W2 to 3 = P2 V = P2 (0 )              = 0 J W3 to 4 = P3 V = P3 ( V3 - V2 )             =  (-3.7014) J W4 to 1 = P4 V = P4 ( 0 )               = 0 J     PART C:

Net work done by the Oxygen W = W1 to 2 + W2 to 3 + W3 to 4 + W4 to 1                                                                                                         = 4.46 J + 0 J + (-3.7014) J + 0 J                                                                                                             = 0.7586 J PART D: Efficiency e = Qout / Qin                                e = ( 2369.775 / 2379.533 )                                 e = 0.995                                e = ( 2369.775 / 2379.533 )                                 e = 0.995                                                                                                                       

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.