$(a)\;-5\;J\qquad(b)\;-10\;J\qquad(c)\;-15\;J\qquad(d)\;-20\;J$

Answer : $\;-5\;J$

Explanation :

In the cyclic process : $\; \bigtriangleup =0$

$\bigtriangleup Q = \bigtriangleup W$

$\bigtriangleup W = W_{A \to B} + W_{B \to C} + W_{C \to A}$

$W_{B \to C} =0 \;$ (Volume remains constant)

$W_{A \to B} = P (V_{2} -V_{1})\;$ (Pressure remains constant)

$ = 10(2-1) = 10\;J$

Therefore , $5 = 10 + W_{C \to A}$

$W_{C \to A} = - 5 \;J$

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