Is Work Done by an Ideal Gas Always Equal to P(Vf - Vi)?

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The work done by an ideal gas during a volume change can be expressed as W = P(Vf - Vi) under constant pressure conditions. This equation simplifies the integral of work done, confirming that it is applicable when pressure remains constant. The discussion clarifies that the work done is indeed path-independent for ideal gases in this scenario. Overall, the relationship holds true as long as the pressure does not vary during the process. The conclusion emphasizes that for constant pressure, the simplified equation accurately represents the work done.
nil1996
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Homework Statement


Workdone by the ideal gas in changing its volume is given by
W=P∫VfdV

Homework Equations


The Attempt at a Solution

My question is can we write the above equation by W=P(Vf-Vi)
Is work done dependent on the path the gas has taken to do work?

The Attempt at a Solution

 
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nil1996 said:

Homework Statement


Workdone by the ideal gas in changing its volume is given by
W=P∫VfdV
That's for constant pressure, right?
 
Yes workdone for constant pressure
 
nil1996 said:
Yes workdone for constant pressure

Then W=P(Vf-Vi) is fine.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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