Two moles of a monatomic ideal gas goes through the cycle represented in Figure below. TA=420...
The figure shows a reversible cycle through which 2.44 mol of a monatomic ideal gas is taken. Assume that p = 2p0lV = 2V0,p0 = 3.91 times 105 Pa, and V0 = 0.0201 m3. Calculate (a) the work done during the cycle, (b) the energy added as heat during stroke abc, and (c) the efficiency of the cycle, (d) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures that occur in the cycle? (e)...
One mole of an ideal monatomic gas is taken through the reversible cycle shown in the figure Volume Process B-C is an adiabatic expansion with PB-13.0 atm and V-4.00x103 m3. The volume at State C is 9.00Vg. Process A-B occurs at constant volume, and Process C A occurs at constant pressure. What is the energy added to the gas as heat for the cycle? Submit Answer Tries 0/10 What is the energy leaving the gas as heat? Submit Answer Tries...
Three moles of an ideal gas are taken around the cycle abc shown in the figure (Figure 1) . For this gas, Cp=29.1 J/(mol?K). Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperatures of the gas in states a, c, and b are Ta=300K, Tc= 492 K, and Tb= 600 K. Calculate the total work W for the cycle
Two moles of a monatomic ideal gas ( CV = 3R/2; CP = 5R/2) are used as the working substance for a heat engine whose cycle is depicted in the Ögure. The path bc is isothermal. In each part below explain the reasoning behind your calculations if necessary. (P0 =1.11 x 105 N/m2; V0 = 4.5 x 10-2 m3) (a) Determine Ta. (b) Determine Tb = Tc. (c) Determine Wab. (d) Determine Qab. (e) Determine Wbc. (f) Determine Qbc. (g)...
0.25 moles ofa monatomic ideal gas starts from point a (400Pa and Im3) in the diagram as shown. It undergoes a constant pressure expansion from a to b (2m3); an isothermal process from b to c (3.2m3); a constant volume process c to d (125Pa); and an isothermal compression from d back to a. Problems 2-5 400 b a 300 2a. Find the temperature values Ta, Tb, Te and Td. 200 100 3 4 1 2 volume (m3) 2b. Find...
Question 4 Incorrect. (a) For 3.07 moles of a monatomic ideal gas taken through the cycle in the figure, where V1 = 4.00V0 what is Wp0Vo as the gas goes from state a to state c along path abc? What is ΔΕί t poVo in going (b) from b to c and (c) through one full cycle? What is AS in o ng (d) from b to c and e through one ful Усе 4 Po Vo す Volume (a)...
A sample (8.1 mol) of an ideal monatomic gas goes through the process shown in Figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat. From C to D,the process is isothermal from D to d,itis sobarie with 3711 of energy leaving the system by heat. P (atm) 12 (m') 0.09 0.2 0.4 (a) Determine the difference in internal energy Eint,B-EinA (b) Determine...
102) 2.37 moles of an ideal monatomic gas initially at 255 K undergoes this cycle: It is (1) heated at constant pressure to 655 K, (2) then allowed to cool at constant volume until its temperature returns to its initial value, (3) then compressed isothermally to its initial state. Find: a. the net energy transferred as heat to the gas (excluding the energy transferred as heat out of the gas). b. the net work done by the gas for the...
A heat engine takes for 0.40 mol of ideal H2 gas around the cycle shown in the pV- diagram.Ta=400KTb=800KTc=592K Process a→b is at constant volume, process b→c is adiabatic, and process c-> a is at constant pressure of 2 atm. The value of y for this gas is 1.40. (a) Find the pressure and volume at points a, b and c (b) Calculate Q, W, and AU for each of the processes. (c) Find the net work done by the gas in the cycle (d)...
A closed system of 2 moles of helium gas ( monatomic) (M=4 g/mol) goes through the following reversible process: Process1->2: Isothermic expansion at T=400k form V1=.03 m^3 to V2=.12 m^3 Process 2->3: Adiabatic expansion to T= 325 K and V3 = .2 m^3 Process 3->4: Isothermic compression at T=325K to P4 = 1.1*10^5 Pa Process 4->1: Adiabatic compression (from state 4 back to state 1) A) sketch a PV diagram of tis cycle B) sketch a ST diagram of this...