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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

Buck Thorn
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puma
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    You expanding for delta U is wrong. For constant p, W = - p. Delta V. – Poutnik May 03 '21 at 16:49
  • @Poutnik I have taken care of that in the very next line. – puma May 03 '21 at 17:04
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    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. – Poutnik May 03 '21 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 04 '21 at 02:25
  • For more on the pV sign convention see https://chemistry.stackexchange.com/questions/66088/why-do-chemistry-and-physics-have-different-sign-convention-in-thermodynamics or https://chemistry.stackexchange.com/questions/139100/what-is-the-correct-expression-of-pressure-volume-work?noredirect=1&lq=1 For more on enthalpy definition as relates to heat at constant p see eg https://chemistry.stackexchange.com/questions/39988/does-enthalpy-equal-heat-when-pv-work-is-done?rq=1 – Buck Thorn May 04 '21 at 04:18
  • See also https://en.wikipedia.org/wiki/Work_(thermodynamics)#Sign_convention – Buck Thorn May 04 '21 at 04:50
  • I tried to interpret what you wrote in Eq 1 but you should check this. It's unclear why you write both pV terms with positive signs, which is why you have received many comments regarding the sign convention. Also worth learning how to format math expressions using mathjax, see eg https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Buck Thorn May 04 '21 at 05:52

1 Answers1

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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 ($\Delta V<0$, $w<0$).

Buck Thorn
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    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. – Poutnik May 04 '21 at 06:22
  • @Buck Thorn♦ That was the clarification i was looking for thanks a lot ;) – puma May 04 '21 at 08:55