Question

A 1 2 2 P. A + D B Pc + When a system of gas is taken from state 4 to 2 along path D and A, 120J of heat flows into the system and 100 J of work is done by the system. When the system is returned to state 4 from 2 along the curved path, E. the absolute value of work done by the system is 40 J. How much heat does the system absorb or reject? 4 VO 3 + Vo VB 10 J absorbed 60 ) absorb 10 J reject ( released) C. d. 60 J reject(released) answer not given e.

Answer 1

AQ - AU + Awe OY First law of thesmadymics will give us the equation, simply Q. AU+W This means if a system takes in on absorbs DQ amount of heat, it is used for the change in_Internal energy cau) will also do some work (w). first law the law of conservation of and the system is based on P 71 А - PA 2 E DA VB Pc < 4 3 Vg vi ges is TOOJ: We know I Given that taken from state 4 to 2 along DA- Heat flow into the system, Q = 120J Work done, pdv. Since from 4 to 1, there is no volume change work done from 4to Lis Lexo. 1005 of work is done when it moves from 1 toa. Now, let calculate the internal when taken from 4to I along DA. energy change Q2 AU_+wi Δυ 2 Q-W- ... 20 J

is curved path. E It is न le Now it is said that the system I returned to state 4 from 2 along the given, the value of work done the system 2 403 Since, it's the work done by the system, it's sign is negative. Walang I - 40J Af it work done I the When it moves from 4to ed along DA, the change in internal energy Now, this internal Shey will decrease by the same amount, le موفیل on the system, would be positive was DoJ Now let's calculate, Q = AU + Walory E -20 - 40 Q - 60T The negative sign shows that the heat is rejected by the system go 60J of heat is rejected or released by the systemy_when_it_mzaves ves along.cuwed path E to 4