# Change in Enthalpy when Pressure is constant

Okay so let me start from the beginning.

I know that $$ΔH= ΔU + p ΔV$$ ( at constant pressure).

Expanding $$ΔU$$,

$$ΔH= q + pΔV + pΔV ~~~~~\text{(q at constant pressure)} \tag{1}$$

Next my chemistry book continued to say that in expansion work done by gas is negative, so the PV work terms cancel out, i.e,

$$ΔH= q - p ΔV + p ΔV$$

Therefore, heat at constant pressure is equal to the change in enthalpy.

Doubt:

Now my question is, in (1) what is the difference between both the PV terms ?

Can't the equation turn out to be: $$ΔH= q - p ΔV - p ΔV= q - 2p ΔV$$?

I am a high schooler and I know that I am missing something important, so can someone help me out?

EDIT:

We may write equation (6.1) as $$∆ U = q -p∆V$$ at constant pressure, where $$q$$ is heat absorbed by the system and $$–p∆V$$ represent expansion work done by the system.

This is what is written in my book. That is, work done in expansion is negative.

Reference - class 11 chemistry ncert book

• You expanding for delta U is wrong. For constant p, W = - p. Delta V. May 3, 2021 at 16:49
• @Poutnik I have taken care of that in the very next line.
– puma
May 3, 2021 at 17:04
• Expansion work done by gas is not negative. But in U and H context, positive work is the one done to the system, not by the system. See also 2 historically opposite notations, Delta U = Q + W versus Q - W. May 3, 2021 at 17:43
• According to the convention in chemistry, isn't the work done BY the gas ON the surroundings (BY the system) negative ? i.e the gas is expanding.....but anyway how does this answer my question ? basically all that I am asking is the physical intuitive difference between the pv work for which enthalpy is defined and the pv work of internal energy. I will check with the convention.
– puma
May 4, 2021 at 2:25
• For more on the pV sign convention see chemistry.stackexchange.com/questions/66088/… or chemistry.stackexchange.com/questions/139100/… For more on enthalpy definition as relates to heat at constant p see eg chemistry.stackexchange.com/questions/39988/… May 4, 2021 at 4:18

Your book uses the "chemistry sign convention", that is, $$w=-p_\textrm{ex} \Delta V$$, where $$p_\textrm{ex}$$ is the external pressure.

However the enthalpy is not defined in terms of the external pressure but rather the pressure $$p$$ in the system: $$H=U+pV$$. From the definition of enthalpy it follows that $$\Delta H= \Delta U+ \Delta (pV)$$, or, at constant system pressure, that $$\Delta H = \Delta U+ p \Delta V = q + \left(-p_\textrm{ex} + p \right) \Delta V$$. Only when the condition of mechanical balance ($$p_\textrm{ex} = p$$) is satisfied at the beginning and end of a process can we say that $$\Delta H = q$$.

For more on the $$pV$$ sign convention see other posts. For more on enthalpy definition as relates to heat at constant pressure see eg this post.

On the sign convention:

If you choose to write $$U=q+w$$ ("chemistry convention") then for constant applied (external) pressure $$w=-p_\textrm{ex} \Delta V$$. Then the sign of the work is regarded as negative when the system expands, doing work on the surroundings ($$\Delta V>0$$, $$w<0$$) or positive when the surroundings does work on the system ($$\Delta V<0$$, $$w>0$$).

If you choose to write $$U=q-w$$ ("physics convention") then for constant applied (external) pressure $$w=p_\textrm{ex} \Delta V$$. Then the sign of the work is regarded as positive when the system expands, doing work on the surroundings ($$\Delta V>0$$, $$w>0$$) or negative when the surroundings does work on the system ($$\Delta V<0$$, $$w<0$$).

• Note that these days physics becomes accepting the chemistry convention as well, so it becomes rather scientific (+) versus engineering (-) convention. All that may create a confusion mess, so it is adviced always making oneself sure which one is used. May 4, 2021 at 6:22
• @Buck Thorn♦ That was the clarification i was looking for thanks a lot ;)
– puma
May 4, 2021 at 8:55