# First Law of Thermodynamics

1. Jan 5, 2014

### Sixty3

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

2. Relevant equations
PV=nRT
dU=dQ+W

3. 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).

2. Jan 5, 2014

### TSny

Hello, sixty3.

Your work looks good to me.

3. Jan 6, 2014

### tomothy

Looks good. You may want to add why dU=0, becuase 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.