An engine takes 325 mole of an ideal monatomic He gas throug

An engine takes 3.25 mole of an ideal monatomic He gas through the cycle shown in the figure. Note that the temperature of the gas does not change during process c-a. Find the pressure at point a. Find the temperature at points a, b, and c. How much heat went in or out of the gas during segments a-b, b-c, and c-a? In each case explain whether heat entered or left the gas.

Solution

Here an Isothermal process [T remains constant], Studying the graph

At point C ; V=0.040m³, P= 2*10⁵ pa, n=3.5 moles

Using the ideal gas law, PV= nRT

we get T= pV/nR= 2*10⁵ *0.040 /3.5 * 8.314= 275 K

a] Pressure at point a is P_a= nRT/ V_a= 3.5 * 8.314*275 K / 0.010= 8*10⁵ pa

b] Since it is an Isothermal process, Temperature remains constant throughout the process. Hence tempertaure at point a, b and c is Ta=Tb=Tc= 275 K

c] In segment a- b

W_ab= [PV]_a- [PV]_b= (8*10⁵*0.01) - (8*10⁵*0.04) = 8000-32000= -24000 J [heat flows outward]

In segment b- c

W_bc= [PV]_b- [PV]_c= (8*10⁵*0.04) - (2*10⁵*0.04) = 32000-8000= 24000 J [heat flows inward]

In segment c- a

W_ca= [PV]_c- [PV]_a= (2*10⁵*0.04) - (8*10⁵*0.01) = 8000-8000= 0 J [ No heat flow]