The work done in Joules when $1.0$ mole of $N_{2} H_{4}$ decomposes against a pressure of 1.0 atm at $27 °$ C is __________
$3 N_{2} H_{4} (l) \to 4 N H_{3} (g) + N_{2} (g)$

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Detailed Solution

$3 N_{2} H_{4} (l) \to 4 N H_{3 (g)} + 2 N_{2 (g)} 1 m o l N_{2} H_{4 (l)} w o u l d p r o d u c e 2 m o l g a s e s$

$(3 m o l N_{2} H_{4} p r o d u c e s 6 m o l o f g a s e o u s p r o d u c t s) P = 1.0 a t m, T = 27 ° C = 300 K$

Work done by system

$= P Δ V = R T Δ n = (8.314 J / m o l / K) (300 K) (2 m o l) = 4988.4 J$

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