A gas is taken through a cyclic process. The change in internal energy along the path from c to a is –160 J. Heat transferred along the path from a to b is 200 J and 40 J for path b to c. Then work done in the cycle is less than 80 J work done in path abc is 80 J work done in path ab is 160 J heat transferred in the system is less than 80 J

Question

Luis Wiggins · Accepted Answer

Q          n      e      t        =      W          n      e      t              U          c      y      c      l      e        =  0        Q          a      b        +      Q          b      c        +      Q          c      a        =      W          a      b        +      W          b      c        +      W          c      a          200  J  +  40  J  +      Q          c      a        =      W          a      b        +      W          b      c        +      W          c      a          240  J  +      Q          c      a        =      W          a      b      c        +      W          c      a          240  J  +      Q          c      a        −      W          c      a        =      W          a      b      c          240  −  160  =      W          a      b      c              W          b      c        =  0    ⇒      Δ        U          b      c        =  40    ⇒      W          a      b        =  80So, answer is heat transferred in the system is less than 80 J.

Answered question

Answer & Explanation

New Questions in Advanced Physics