Isothermal exapansion, work done

[Saxby]

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 :)

 

[sankalpmittal]

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.

 

[Saxby]

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.

 

[sankalpmittal]

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.

 

[Nickie]

To derive an expression for the amount of work done during an isothermal expansion of an ideal gas, we can use the ideal gas law, PV = nRT, and the definition of work done, W = Fd. In this case, the force F is the pressure of the gas, P, and the distance d is the change in volume, V2-V1.

Starting with the ideal gas law, we can rearrange it to solve for pressure: P = (nRT)/V. Substituting this into the definition of work, we get:

W = (nRT/V)d

Since the temperature T is constant during an isothermal expansion, we can take it out of the integral. Also, since the gas is expanding, dV is positive. Therefore, we can rewrite the integral as:

W = nRT∫V1V2(1/V)dV

Integrating this expression, we get:

W = nRT[ln(V2) - ln(V1)]

Simplifying further, we get:

W = nRTln(V2/V1)

This is the expression for the amount of work done during an isothermal expansion of one mole of an ideal gas. It shows that the work done is directly proportional to the number of moles of gas (n), the gas constant (R), and the temperature (T). It is also proportional to the natural logarithm of the ratio of final and initial volumes. This makes sense as the work done increases as the gas expands and the final volume becomes larger than the initial volume.