Why is P_ext different for reversible and irreversible isothermal expansion?

[AdityaDev]

I was able to derive the work done in a reversible isothermal expansion. There as the P changes to P-dP, the volume increases by dV and hence the internal pressure also decreases by dP and equilibrium is maintained. (Thanks to @Nugatory and @Chestermiller fo r explaining it).
Now when we take an irreversible isothermal expansion in the book the $P_{ext}$ is taken constant and then dV is integrated to get
$$W = -P_{ext}(V_2-V_1) $$
But for reversible isothermal expansion, the $P_ext$ is taken as a variable. That is it is taken as $\frac{nRT}{V}$ and then integrated to get
$$W = RTln(\frac{V_2}{V_1})$$
why is $P_{ext}$ different for the two cases?

 

[Raghav Gupta]

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.
If $p_{int}>p_{ext}$ in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process.

On the other hand if $p_{int}=p_{ext}$ then the process is reversible.
In that case $p_{int}=p_{ext}=p$
You have written the other way round.
That that in reversible process the p is variable whereas in irreversible constant.
Can you check on that?

 

[Chestermiller]

The equation you wrote for an irreversible expansion assumes a constant force per unit area Pext applied to the gas, and also, there should be a plus sign rather than a minus sign.

In an irreversible expansion, viscous stresses can contribute to Pext and, in addition, the pressure and/or the temperature within the gas are not uniform spatially, so we can't use the ideal gas law to determine the force per unit area exerted by the gas on the boundary (Pext). So the ideal gas law cannot be used to calculate the work done on the surroundings.

For both reversible and irreversible processes, the work done on the surroundings is equal to the integral of Pext dV. For an irreversible process, Pext is not equal to the pressure calculated from the ideal gas law P, but for a reversible process, the deformation is slow (such that viscous stresses are negligible), and also Pext = P = nRT/V. For an irreversible process, we can control Pext as a function of time t during the deformation to be whatever we wish.

Chet

 

[AdityaDev]

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.
If $p_{int}>p_{ext}$ in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process.

On the other hand if $p_{int}=p_{ext}$ then the process is reversible.
In that case $p_{int}=p_{ext}=p$
You have written the other way round.
That that in reversible process the p is variable whereas in irreversible constant.
Can you check on that?

The expressions are correct

 

[AdityaDev]

@Raghav Gupta ,take a look at @Chestermiller 's reply. The first paragraph solved my doubt. Second paragraph explains the reason why you can't take P=nRT/V for irreversible process.

 

[Raghav Gupta]

@Raghav Gupta ,take a look at @Chestermiller 's reply. The first paragraph solved my doubt. Second paragraph explains the reason why you can't take P=nRT/V for irreversible process.

I was also a bit confused.Maybe I have read a wrong article or not clearly understood.Really these mentor guys explain in a nice detailed way.