Heat and Thermodynamics 5 Question 32

33. 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.

(2011)

Column I Column II
(A) Process $A \rightarrow B$ (p) Internal energy
decreases
(B) Process $B \rightarrow C$ (q) Internal energy
increases
(C) Process $C \rightarrow D$ (r) Heat is lost
(D) Process $D \rightarrow A$ (s) Heat is gained
(t) Work is done on the gas
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Answer:

Correct Answer: 33. (A) $\rightarrow$ p, r, t; (B) $\rightarrow$ p, r; (C) $\rightarrow$ q, s; (D) $\rightarrow$ r, t

Solution:

  1. Internal energy $\propto T \propto p V$

This is because $U=\frac{n f}{2} R T=\frac{f}{2}(p V)$

Here, $n=$ number of moles, $f=$ degree of freedom.

$\therefore$ 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,

$$ Q=W+\Delta U $$

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 $\Delta U$ is positive if product of $p V$ is increasing and $\Delta U$ is negative if product of $p V$ 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) $\rightarrow(p, r)$; $(C) \rightarrow(q, s) ;(D) \rightarrow(r, t)$



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