$(a)\;15\;J\qquad(b)\;11\;J\qquad(c)\;10\;J\qquad(d)\;14\;J$

Thermodynamics

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Answer : $\;14\;J$

Explanation :

Work done at constant pressure

$=P \bigtriangleup V=P\;(V_{2}-V_{1}) = 4 J$

$PV=nRT$

$PV_{1}=nRT_{1}----(1)$

$P(V_{2}-V_{1})=nR(T_{2}-T_{1})=4$

$PV_{2}nRT_{2}-----(2)$

$\bigtriangleup u= \large\frac{nR}{(r-1)}\;(T_{2}-T_{1})=\large\frac{4}{r-1}$

For a diamatic gas , $\;r=\large\frac{7}{5}$

Therefore , $\;\bigtriangleup u =\large\frac{4}{\large\frac{7}{5}-1}=5 \times 2=10 J$

$\bigtriangleup Q = \bigtriangleup u + \bigtriangleup w$

$=10+4 = 14 J$

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