$(a)\;7 : 5 :2\qquad(b)\; 7 : 2 :5\qquad(c)\;3 : 5 : 2\qquad(d)\;3 : 2 :5$

Thermodynamics

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Answer : $3 : 2 :5$

Explanation :

At constant pressure

$\bigtriangleup Q= n C_{v} \bigtriangleup T \;, \bigtriangleup u = n C_{v} \bigtriangleup T$

Therefore , $\;\bigtriangleup W = n \bigtriangleup T (C_{p}-C_{v})= nR \bigtriangleup T$

For a mono atomic gas ,

$C_{p}= \large\frac{5R}{2} \quad \; C_{v} = \large\frac{3R}{2}$

Therefore , $\; \bigtriangleup u : \bigtriangleup W : \bigtriangleup Q $

$= C_{v} : R : C_{p}$

$=\large\frac{3}{2} : 1 : \large\frac{5}{2} = 3 : 2 : 5$

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