# Thermodynamics, Isothermal Question

1. Aug 24, 2010

### kalbuskj31

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

10 moles of an ideal gas, in the initial state P1 = 10 atm, T1 = 300K are taken around the following cycle.

a) A reversible change of state along a straight line path on the P-V diagram to the state P = 1 atm T = 300K. How much work is done during the cycle?

2. Relevant equations

Isothermal, ΔU = 0 so w = p dV

3. The attempt at a solution

w = p dV and p = nRT/V so w = nRT (dV/V)

w = nRT ln(Vf/Vi) = nRT ln(Pi/Pf)

w = [(10 moles)* (8.3144 J/ K *mole) * (300K)] ln(10atm/1atm)

w = 57.4 kJ but the correct answer is 123.4kJ. Suggestions??

2. Aug 25, 2010

### Andrew Mason

The process is not isothermal. You are not taking into account that the graph is a straight line.

You are finding the area under the isothermal curve between the beginning and ending points (PV = constant).

AM

3. Aug 25, 2010

### kalbuskj31

Thanks. Can you please explain why this process isn't considered isothermal?

4. Aug 25, 2010

### presbyope

An isotherm has the same temperature during the whole process, not just at the start and end points. It will look like an exponential curve on the PV diagram. Adiabats will be "steeper" because they cross the isotherms.

Since it's a straight line, did you try a graphical integration?

5. Aug 25, 2010

### kalbuskj31

Thank you for the help. the graphical interpretation helped give me the correct answer. I've been out of school for 3 years, so I'm brushing up on everything.