A sample of an ideal gas goes through the process shown in the figure below. Not

Question

A sample of an ideal gas goes through the process shown in the figure below. Notice that pressure is given in atmospheres. 1 atm = 101300 Pa.

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 A, it is isobaric with 371 kJ of energy leaving the system by heat.

a. Determine the difference in internal energy from B - C.

b. What is the change in internal energy from C to D?

c. What is the change in internal energy from D to A?

d. What is the change in internal energy from A to B?

blackboard.und.edu Take Test Homework-PHYS 251: University Take Test: Chapter 20 Homework- PHYS 251: University Question Completion Status A sample of an ideal gas goes through the process shown in the figure below. Notice that pressure is given in atmospheres. 1 atm 101300 Pa. From A to B, the process is adiabatic from B to C, it is isobaric with 345 k of energy entering the system by heat; from C to D, the process is isothermal; from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. a. Determine the difference in internal energy from B-C P (atm) 0.09 0.2 0.4 250791 1 points Save Answer QUESTION 3 A sample of an ideal gas goes through the process shown in the figure above From A to B, the process is adiabatic rom 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 A, it is isobaric with 371 kJ of energy leaving the system by heat b. What is the change in internal energy from C to D? Saxe All Answers Save and Submit Click Save and Submit to save and submit. Cick Sove All Answers to save olf answers

Explanation / Answer

Given,

Considering the given PV diagram,

Eint,B - EintA = (Eint,B - EintC) + (Eint,C - Eint,D) + (Eint,D - EintA)
As the process from C to D is siothermal, Eint,C - EintD = 0
=> Eint,B - EintA = (Eint,B - EintC) + (Eint,D - EintA) ... ( 1 )


For the process BC,
Q = EintC - Eint,B + W
=> 345 = EintC - Eint,B + PdV
=> 345 = EintC - Eint,B + 3 * (0.40 - 0.09) * (101.325) ... [1atm = 101.325 kN/m^2]
=> Eint,B - EintC = - 250.76775 kJ ... ( 2 )

For the process DA,
Q = EintA - Eint,D + W
=> - 371 = EintA - Eint,D + PdV
=> - 371 = EintA - Eint,D + 1 * (1.20 - 0.20) * (101.325) ... [1atm = 101.325 kN/m^2]
=> Eint,D - EintA = 269.675 kJ ... ( 3 )

Plugging from ( 2 ) and ( 3 ) in ( 1 ),
Eint,B - EintA
= - 250.76775 kJ + 269.675 kJ
= 18.90725 kJ.

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