Isothermal exapansion, work done

AI Thread Summary
The discussion focuses on deriving the work done during the isothermal expansion of one mole of an ideal gas from volume V1 to V2 at a constant temperature T. Participants clarify that the process should be considered reversible to calculate the maximum work done by the gas. The equation W = V2∫V1Pdv is highlighted, with the ideal gas law (PV = nRT) used to express pressure in terms of volume. A hint is provided to replace pressure in the work equation, leading to the final expression W = RT * ln(V2/V1). The importance of careful integration is emphasized to arrive at the correct result.
Saxby
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Homework Statement


Derive an expression for the amount of work done when one mole of an ideal gas expands isothermally at a temperature T from an initial volume V1 to a final volume V2.


Homework Equations


PV = nRT
W = V2∫V1Pdv

The Attempt at a Solution


I'm not really sure how to go about this, I know if it's isothermal the temperature doesn't change so Volume would be inversely proportional to Pressure. But other than that i don't know what to do, to be honest I'm not really exactly what they want me to write. Any help would be much appreciated :)
 
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Saxby said:

Homework Statement


Derive an expression for the amount of work done when one mole of an ideal gas expands isothermally at a temperature T from an initial volume V1 to a final volume V2.


Homework Equations


PV = nRT
W = V2∫V1Pdv

The Attempt at a Solution


I'm not really sure how to go about this, I know if it's isothermal the temperature doesn't change so Volume would be inversely proportional to Pressure. But other than that i don't know what to do, to be honest I'm not really exactly what they want me to write. Any help would be much appreciated :)

The question seems to be incomplete. It is isothermal expansion , fine. But its isothermal reversible or isothermal irreversible process , its just not mentioned.

I ask you a question : Do you want to calculate maximum work done by the gas , or not in an isothermal process. If former , consider the process reversible , else irreversible.

Oh , they do not concern you about external pressure. Consider your process reversible then.

Hint :

W = V2∫V1Pdv ...(i)

PV=nRT => P=nRT/V

Now replace this value of P in (i) and carry on your integration.
 
Thanks for the Hint that's makes sense, my final answer (upon using that hint) would be W = RT * Ln(V). I think that's the answer but i'll try and confirm it with my lecturer tommorow if i can. Thanks for your help.
 
Saxby said:
Thanks for the Hint that's makes sense, my final answer (upon using that hint) would be W = RT * Ln(V). I think that's the answer but i'll try and confirm it with my lecturer tommorow if i can. Thanks for your help.

No you should get W=RT*ln(V2/V1)

See your integral again. Be careful while doing integration.
 
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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