Constant pressure

[Rosa Munoz 2E]

When they refer to constant pressure, are we referring to constant external pressure?

[Sahil Jog 1F]

Yes they are referring to constant external pressure, since a dynamic external pressure would do unnecessary work on a system, and thus be unreasonable problem to solve.

[faithkim1L]

Constant external pressure.

[Vinita Saxena 2I]

Yes, constant external pressure

[Callum Guo 1H]

Constant external pressure

[205389184]

Yes, you are correct, we are referring to constant pressure

[Jacob Villar 2C]

Constant external pressure.

[Alexis Robles 2k]

yeah, its Constant external pressure.

[rohun2H]

Yes, constant external pressure.

[705367472]

yes, we are referring to constant external pressure

[Matt Sanruk 2H]

And this would be irreversible?

[Adriana_4F]

And this would be irreversible?

Yes this would be irreversible

[Radha Patel 4I]

It would be irreversible and it would be referring to external pressure so w=0

[aishwarya_atmakuri]

Yes, they would be referring to a constant external pressure.

[CameronDis2K]

Yes, constant external pressure, as it allows the system to change somewhat independent of the external environment. For example, in the equation for an irreversible expansion work, W = - P external * delta V. However, if it is a gas in a container, pressure of the gas may be determined through the ideal gas law (but the system pressure will change).

[KBELTRAMI_1E]

Is it always assumed to be 1 atm? What about in a vacuum?

[Vinita Saxena 2I]

In a vacuum there is no external pressure, so there is nothing to push back against, meaning w=0

[Micah3J]

And so just because there is constant external pressure, the internal pressure can still be changing because of the reaction going on?

[RobertXu_2J]

Yes, constant external pressure.

[Mulin_Li_2J]

Yes and I can give you some explanation on this.

You might refer to "constant pressure" mentioned when we try to calculate the amount of work done by a system. First, let's take a look at the definition of work: w = F(o)*D. F(o) here refers to an opposing force. Basically, this formula implies that you have to have an opposing force to do work, and the work you have done is how far you got by pushing against the opposing force. Hence, because w = -P*deltaV is a derived formula from w = F(o)*D, P also refers to external pressure.

Hope this can help! -_+