# Work done by system of ideal gas in isobaric expansion

Consider a system of ideal gas in a container with piston and the isobaric expansion of gas takes place.
As the process is isobaric, so initially the pressure of gas is equal to atmospheric pressure as the piston remains in equilibrium over the container. Then, first we increase the temperature and as a result pressure of gas increases and the piston rises till the pressure reduces to atmospheric pressure. So at initial and final state pressure of gas is constant thus the process is isobaric.
So, work done by gas should be
$$dW=F_{applied\;by\;gas}(x).dx$$
$$dW=P_{applied\;by\;gas}(x)Adx$$
$$s.t.\;P_{applied\;by\;gas}(x_{initial\;state})=P_{applied\;by\;gas}(x_{final\;state})=P_{atm}$$
So, total work done, $$W=\int_{x_{initial}}^{x_{final}}P(x)Adx$$
So, we should know pressure as a function of $$x$$ to calculate the work done by gas.

My question is that in books, they directly use the argument that work done by gas is dependent on $$P_{external}$$ and as it is constant, they give the expression of work as $$dW=-P_{external}Adx$$.
I don't understand how they do that and use external pressure in the expression of work done.
I think we should take into account pressure of gas as a function of $$x$$ to find the work done by gas.

• Isobaric means with constant external pressure. Then it depends if the process is reversible ( always considered in equilibrium ) or not. For the former, internal and external pressures are equal and constant. For more, see Reversible process on wikipedia. Commented Dec 9, 2020 at 11:09
• Consider that for reversible isobaric expansion $p_\mathrm{int} = p_\mathrm{ext} + \mathrm{d}p$ Commented Dec 9, 2020 at 11:18
• Thanks for the reply. But if the process is irreversible then I think we have to take into account the pressure of gas as a function of x.
– Manu
Commented Dec 9, 2020 at 11:19
• Because then $p_{int}=p_{ext}+\Delta p$ and $\Delta p$ can't be neglected
– Manu
Commented Dec 9, 2020 at 11:21
• For example at initial and final state, pressure of gas is same as external pressure andbetween the states it varies as explained in my question. But, in high school chemistry book, it is written that work done for doth reversible and irreversible process is -$P_{ex}\Delta V$. I don't understand that.
– Manu
Commented Dec 9, 2020 at 11:26