# Work Done by Isothermal Expansion

1. Dec 6, 2014

### Calu

1. The problem statement, all variables and given/known data
A quantity of ideal gas (0.800mole) at a pressure of 10.0atm and 200K is allowed to
expand isothermally until it reaches a pressure of 1.00atm. Calculate the work done
if this expansion is carried out a) against a vacuum, b) against a constant external
pressure of 1.0atm and c) reversibly.

I'm not even sure where to start here. I've only calculated work done using dW = ∫ PdV. However, I'm not given an initial or final volume here, so I'm not sure how to proceed. Any hints?

2. Dec 6, 2014

### ehild

It is ideal gas. Given the amount of gas, the pressure and the temperature. What is the volume?

3. Dec 6, 2014

### Calu

I see, I simply rearrange PV = nRT?

In which case, how does my calculation change between each of the above instances?

4. Dec 6, 2014

### ehild

I do not see any calculations of yours. :)

5. Dec 6, 2014

### Calu

How would I go about calculating work when both the pressure and the volume change? I can integrate P.dV and put in the values for the initial and final volumes, but what value of pressure do I use?

6. Dec 6, 2014

### ehild

Pressure is only defined for reversible (quasi-static) process. In that case, you can express P from the ideal gas law.

In the other cases, you can calculate the work done on the gas by external forces.
What is the work done on the gas if there is no external force (when the gas expands into vacuum)?
What is the work done on the gas if the external pressure is constant (1.0 atm)?

7. Dec 6, 2014

### Calu

Okay, so in the reversible case I can substitute nRT ∫ 1/V .dV = nRT ln(V2-V1). I'm not sure how to work out the other two cases though.

8. Dec 6, 2014

### ehild

What is the work done if no force is exerted?
What is the work done by the environment on the gas if the external pressure is 1.0 atm and the volume changes from V1 to V2?

9. Dec 6, 2014

### Calu

Okay, so in the first case, I would say no work is being done on the gas, however the question just asks for the work done. Is the gas doing work on the surroundings in this case? In the second case I would say that the numerical value of work done is equivalent to the difference between the volumes.

10. Dec 6, 2014

### ehild

Is there anything to do work on if the gas expands freely?
It is the constant external pressure multiplied by the difference of volumes, and the external pressure is 1.0 atm,

Last edited: Dec 6, 2014
11. Dec 6, 2014

### Calu

In first case, I guess not. That makes sense. In the second case I apologise for omitting the multiplication by 1, its how I've been taught. I very really show multiplication by 1 in my working.

12. Dec 6, 2014

### ehild

It is not multiplication by 1 but multiplication by 1.0 atm. And the result should be in joules. So what is the numerical value at the end?