One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E &#160…

One mole of a monatomic ideal gas is taken along two cyclic processes E F G E  and  E F H E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.
List I List II A. G E   P . 1 6 P 0 V 0  ln 2 B. G H   Q . 3 6  P 0 V 0 C. F H   R . 2 4  P 0 V 0 D. F G   S . 3 1  P 0 V 0

  1. a-q;b-s;c-r;d-p;
  2. a-s;b-r;c-q;d-p;
  3. a-q;b-s;c-r;d-p;
  4. a-q;b-r;c-s;d-p;

Solution


F G  work done in isothermal process is nRT ln V f V i = 3 2 P 0 V 0 ln 3 2 V 0 V 0
= 3 2 P 0 V 0 ln 2 5 = 1 0  P 0 V 0 ln 2
ln  G E , Δ W = P 0 3 1 V 0 = 3 1  P 0 V 0
ln  G H  work done is less than 3 1  P 0 V 0  i.e., 2 4 P 0 V 0
ln  F H  work done is 3 6  P 0 V 0 :

Asked in: JEE Advanced 2013 (Paper 2)

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