# Thermodynamics: Calculating the work done

1. Jun 16, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
Work done in isothermal process is
$$nRT\ln\frac{V_2}{V_1}$$
In the given question, $n=1 \text{mol}$ and $V_2=nV_1$. Hence work done is:
$$RT\ln n$$
but this is wrong.

2. Jun 16, 2013

### I like Serena

Hey Pranav!

You have calculated the work done on the gas, but you have omitted the minus sign. :surprised
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.

3. Jun 16, 2013

### voko

Where is the atmosphere in your equations?

4. Jun 16, 2013

### Pranav-Arora

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?

5. Jun 16, 2013

### voko

This looks good to me.

6. Jun 16, 2013

### I like Serena

Yep. Looks good.

7. Jun 16, 2013

Thanks!