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Question: An ideal gas expands by performing 200 J of work, during this internal energy increases by 432 J. Wh...

An ideal gas expands by performing 200 J of work, during this internal energy increases by 432 J. What is enthalpy change?

A

200 J

B

232 J

C

432 J

D

632 J

Answer

632 J

Explanation

Solution

Given:

  • Work done by the gas, W=200 JW = 200 \, \text{J} (expansion work done by the system).
  • Increase in internal energy, ΔU=432 J\Delta U = 432 \, \text{J}.

Using the First Law of Thermodynamics:

Q=ΔU+W=432+200=632 JQ = \Delta U + W = 432 + 200 = 632 \, \text{J}

For a process involving only PdVPdV work (constant pressure), the enthalpy change is given by:

ΔH=Q=632 J\Delta H = Q = 632 \, \text{J}