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Question
Answer the following question.
Calculate the work done during the synthesis of NH3 in which volume changes from 8.0 dm3 to 4.0 dm3 at a constant external pressure of 43 bar. In what direction the work-energy flows?
Solution
Given:
Initial volume (V1) = 8.0 dm 3
Final volume (V2) = 4.0 dm 3
External pressure (Pext) = 43 bar
To find:
The work done (W) and direction of the work energy flow.
Formulae: W = - Pext Δ V = - Pext (V2 - V1)
Calculations:
From formula,
W = - Pext Δ V = - Pext (V2 - V1)
∴ W = - 43 bar × (4.0 dm3 - 8.0 dm3) = 172 dm3 bar
Now, 1 dm3 bar = 100 J
Hence, 172 dm3 × `(100 "J")/(1 "dm"^3 "bar")` = 17200 J = 17.2 kJ
Since, the work is done on the system, work-energy flows into the system from surroundings.
∴ The work done (W) = 17.2 kJ
∴ Work energy flows into the system.
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