Thermodynamics: Calculating the work done

[Saitama]

Homework Statement

One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature ##T##. The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally the gas volume under the piston ##n## times by slowly raising the piston? The friction of piston against the cylinder walls is negligibly small.

Homework Equations

The Attempt at a Solution

Work done in isothermal process is
[tex]nRT\ln\frac{V_2}{V_1}[/tex]
In the given question, ##n=1 \text{mol}## and ##V_2=nV_1##. Hence work done is:
[tex]RT\ln n[/tex]
but this is wrong.

 

[I like Serena]

Hey Pranav!

You have calculated the work done on the gas, but you have omitted the minus sign.
So if you (slowly) extract an amount of energy equal to ##RT\ln n## from the gas, the process will be isothermal.
This is not the total amount of work.

Note that it is a bit hard to force a change and still getting energy back instead of putting it into it.

 

[voko]

Where is the atmosphere in your equations?

 

[Saitama]

Where is the atmosphere in your equations?

Hey Pranav!

You have calculated the work done on the gas, but you have omitted the minus sign.
So if you (slowly) extract an amount of energy equal to ##RT\ln n## from the gas, the process will be isothermal.
This is not the total amount of work.

Note that it is a bit hard to force a change and still getting energy back instead of putting it into it.

Thanks voko and ILS for the replies!

Okay, so there will be some work done on the atmosphere too. The change in volume of atmosphere is ##(n-1)V_1##. The work done on the atmosphere is ##PV_1(n-1)## where P is the initial pressure. Also ##PV_1=RT##, hence net work done is ##(n-1)RT-RT\ln n=RT(n-1-\ln n)##, correct?

 

[voko]

This looks good to me.

 

[I like Serena]

Yep. Looks good.

 

[Saitama]

Thanks!