How Is Work Calculated in an Isothermal Expansion of an Ideal Gas?

[Sixty3]

Homework Statement

5 litres of an ideal gas which is initially at 300K and 10atm, and is expanded to a final pressure of 1atm. Find work done, change in internal energy, heat absorbed and change in enthalpy of the system if the process is isothermal and reversible.

Homework Equations

PV=nRT
dU=dQ+W

The Attempt at a Solution

I just want to check if I have the correct method here.
So using PiVi=nRT to find the number of moles, then the work done is -nRTln(Vf/Vi), where Vf/Vi is the ratio Pi/Pf. Because it is an isothermal process dU=0 → dQ=-W. Also the change in enthalpy will be zero.

Thank you for any help.

(I have posted this here because there is a second part to the question which I may need help for, but I just want to check if I have the right idea).

 

[TSny]

Hello, sixty3.

Your work looks good to me.

 

[tomothy]

Looks good. You may want to add why dU=0, because for an isothermal process, in general dU is not necessarily zero. It is true because dU, for a system which changes its volume by dV and temperature by dT, dU is also given by:

\textrm{d}U=C_V \textrm{d}T + \left ( \frac{\partial U}{\partial V} \right )_T \textrm{d}V

And you also know that for a perfect gas, U(T)=U(0)+C_V T, the internal energy for a perfect gas has no volume dependence because there are no intermolecular interactions.