# Isothermal exapansion, work done

• Saxby
In summary: Remember , you can't integrate P directly here , you need to use P=nRT/V and then integrate.In summary, to calculate the 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, we use the equation W = V2∫V1Pdv, where P=nRT/V. The final answer is W=RT*ln(V2/V1).
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.

PV = nRT
W = V2V1Pdv

## 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 :)

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.

PV = nRT
W = V2V1Pdv

## 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 = V2V1Pdv ...(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.

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.

## 1. What is isothermal expansion?

Isothermal expansion is a thermodynamic process in which a system expands or contracts at a constant temperature. This means that the energy of the system remains constant throughout the process.

## 2. How is work done in an isothermal expansion?

Work is done in an isothermal expansion by the system as it expands against external pressure. This work is done at a constant temperature, so the change in internal energy of the system is equal to the work done.

## 3. What is the formula for calculating work done in an isothermal expansion?

The formula for calculating work done in an isothermal expansion is W = nRT ln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature, and V2 and V1 are the final and initial volumes, respectively.

## 4. How does isothermal expansion differ from adiabatic expansion?

Isothermal expansion occurs at a constant temperature, while adiabatic expansion occurs without any heat exchange with the surroundings. This means that the temperature of the system can change in adiabatic expansion, but not in isothermal expansion.

## 5. What are some real-world examples of isothermal expansion?

Some real-world examples of isothermal expansion include the expansion of a balloon as it is heated, the expansion of steam in a steam engine, and the expansion of air in a bicycle pump as it is compressed.

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