Discussion Forum



1)

One mole of a monatomic ideal gas is taken through a cycle ABCDA 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

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A) $(A) \rightarrow p,r,t ; (B) \rightarrow p,t;(C) \rightarrow q.s;(D) \rightarrow r$

B) $(A) \rightarrow p,r,s ; (B) \rightarrow p,r;(C) \rightarrow q.s;(D) \rightarrow r$

C) $(A) \rightarrow p,r,t ; (B) \rightarrow p,r;(C) \rightarrow q.s;(D) \rightarrow p$

D) $(A) \rightarrow p,r,t ; (B) \rightarrow p,r;(C) \rightarrow q.s;(D) \rightarrow r,t$

Answer:

Option D

Explanation:

 Internal energy $\propto T \propto pV$

 This is because 

 $U= \frac{nf}{2}RT=\frac{f}{2}pV$

 Here, n= number of moles

 f= degree of freedom

If the product pV increases, then internal energy will increase and if the product decreases, the internal energy will decrease.

Further, work is done on the gas, if the volume of gas decreases. For heat exchange

 $Q= W+\triangle U$

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  $\triangle U$ is 

positive if product of pV is increasing and $\triangle U$ is negative if the product of pV is decreasing. If heat is taken by the gas Q is positive and if heat is lost by the gas Q is negative.
Keeping the above points in mind the answer to this question is as under.

 (A) $\rightarrow$ (p,r,t)

 (B) $\rightarrow$ (p,r)

 (C) $\rightarrow$ (q,s)

 (D) $\rightarrow$ (r,t)