Understanding the Meaning of dW in Calculus

[doctordiddy]

What is the difference between dW change in work and work?

For example, can you give me a case where you would use both to calculate something?

Thanks.

 

[Vorde]

Well.

'W', is work, also known as the change in potential energy of an object over a distance, which can be represented as $W=FD$, where the force is a constant.

'dW', is an infinitesimal amount of work, meaning its the work when the distance traveled is 'infinitely small'.

Now I couldn't tell you where you would use dW though. Where did you see this?

 

[doctordiddy]

Well.

'W', is work, also known as the change in potential energy of an object over a distance, which can be represented as $W=FD$, where the force is a constant.

'dW', is an infinitesimal amount of work, meaning its the work when the distance traveled is 'infinitely small'.

Now I couldn't tell you where you would use dW though. Where did you see this?

I got it from the equation dW=PdV

does this mean that dV is a change in volume or an infinitesimal value?

In addition can you tell me if the equations W=PV and W=nRT are valid?

 

[Vorde]

Ah.

Usually when I've seen that equation, its δW = p dV.
The lowercase delta here means 'change', not infinitesimal change (though I've been taught that the lowercase delta means a small change compared to the normal capital delta).

To your second question, yes to both. As you should have been taught in calculus, a differential is an infinitesimally small change in something. So dV is an infinitesimal value, of the change in volume.

Sorry, I cannot help you with your last question, I'm not familar with the equations and I don't want to mislead you.

 

[mikeph]

I got it from the equation dW=PdV

does this mean that dV is a change in volume or an infinitesimal value?

dW and dV are both changes, and are both infinitesimal. The equation is essentially a reformulated version of an expression for the derivative, dW/dV = P. As long as it's understood that dW and dV are linked as numerator and denominator are in a derivative, then the equation should make sense.

In essence it means "the tiny amount of work done is equal to the pressure times the tiny amount of change in volume". And when we say "tiny", we mean infinitesimally small.

W=PV is not something I've ever seen, and it doesn't even make sense. Work done should always be defined relative to a start point and an end point. If you're going to integrate dW = P*dV you need limits on the right side as well as the left. So the equation would look more like: W = P(V2-V1). That is assuming P is constant.

 

[doctordiddy]

dW and dV are both changes, and are both infinitesimal. The equation is essentially a reformulated version of an expression for the derivative, dW/dV = P. As long as it's understood that dW and dV are linked as numerator and denominator are in a derivative, then the equation should make sense.

In essence it means "the tiny amount of work done is equal to the pressure times the tiny amount of change in volume". And when we say "tiny", we mean infinitesimally small.

W=PV is not something I've ever seen, and it doesn't even make sense. Work done should always be defined relative to a start point and an end point. If you're going to integrate dW = P*dV you need limits on the right side as well as the left. So the equation would look more like: W = P(V2-V1). That is assuming P is constant.

according to what you said for dW=PdT, does that mean dW=nRdT as well?

 

[mikeph]

For an isobaric expansion of an ideal gas, yes.

 

[Khashishi]

Infinitesimals only have meaning when compared to other infinitesimals. It tells you how much one value changes when another value changes by a small amount. dW doesn't mean anything by itself, but it can be used in an equation with dT or dV or some combination.