JEE Advanced2011PhysicsThermodynamicsActual
One mole of a monatomic ideal gas is taken through a cycle A B C D A as shown in the p -V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.
Options
- A(A) p,q,r,t, (B) q,r, (C) q,s, (D) r
- B(A) p,q,t, (B) p,r, (C) q,r, (D) s
- C(A) p,s, (B) q,r, (C) q,s, (D) r
- D(A) p,q,r,t, (B) p,q,r, (C) q,r, (D) s
Correct answer
B. (A) p,q,t, (B) p,r, (C) q,r, (D) s
Step-by-step solution
Internal energy T p V This is because U= n f 2 R T= f 2 p V Here, n= number of moles f= degree of freedom If the product p V increases, then internal energy will increase and if product decreases, the internal energy will decrease. Further, work is done on the gas, if volume of gas decreases. For heat exchange. Q=W+ U Work done is area under p-V graph. If volume increases work done by gas is positive and if volume decreases work done by gas is negative. Further U is positive if product of p V is increasing and N U