In a constant-pressure process, [tex]\Delta H = 0[/tex]. What can you conclude about [tex]\Delta E[/tex], [tex]q[/tex], and [tex]w[/tex]?

1. If [tex]\Delta H = 0[/tex], [tex]q_p = w[/tex] and [tex]\Delta E = 0[/tex].
2. If [tex]\Delta H = 0[/tex], [tex]q_p = \Delta E[/tex] and [tex]w = 0[/tex].
3. If [tex]\Delta H = 0[/tex], [tex]q_p = 0[/tex] and [tex]\Delta E = 0[/tex], [tex]w = 0[/tex].
4. If [tex]\Delta H = 0[/tex], [tex]q_p = \Delta E = w[/tex].
5. If [tex]\Delta H = 0[/tex], [tex]q_p = 0[/tex] and [tex]\Delta E = w[/tex].

In a constant-pressure process, ΔH represents the change in enthalpy, which is the heat exchanged at constant pressure. When ΔH = 0, it means that there is no heat exchange with the surroundings.

According to the First Law of Thermodynamics, the change in internal energy (ΔE) is related to heat (q) and work (w) as follows:

ΔE = q - w

If ΔH = 0, it implies that there is no heat exchange (q = 0) at constant pressure. Therefore, we have:

ΔE = q - w

ΔE = 0 - w

ΔE = -w

This means that the change in internal energy (ΔE) is equal to the negative of the work done (w). Consequently, if ΔH = 0, we can conclude that qp (heat at constant pressure) is equal to ΔE, and the work done (w) is zero.

Therefore the correct option is 2

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