When a pressure of 500 Nm-2 applied the volume changes from 0.05m3 to 0.03m3 then work done by the system is:
Correct answer: A. 10 J
- A. 10 J
- B. -5 J
- C. -10 J
- D. +5 J
Explanation
W=PΔV ΔV= Change in volume = -0.03m3-0.05m3=-0.02m3 W=−(500Nm−2)×(−0.02m3)=10J Hence this option is incorrect.
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About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
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