When the first law of thermodynamics, Q = $\Delta$u + W, is applied to an ideal gas that is taken through an isobaric process,

Question

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In an isobaric process (constant pressure), work done (W) is not zero because the volume changes, and the internal energy change ($\Delta$u) is also not zero due to temperature change. Therefore, heat added to the system (Q) is used for both changing the internal energy and doing work, making none of these quantities zero.

When the First Law of Thermodynamics, Q = Δ\DeltaΔ U + W, Is Applied to an Ideal Gas That

When the First Law of Thermodynamics, Q = $\Delta$ U + W, Is Applied to an Ideal Gas That