# Heat and the first Law of thermodynamics

I have the following question and I'm still sortof confused as to how the change in internal energy is zero.

Question: A sample of $1.5\,\mathrm{mol}$ of diatomic gas is at an initial pressure of $3\,\mathrm{atm}$ and at a volume of $2.2\,\mathrm{L}$. the gas expands isothermally until the volume doubles. what is the change in internal energy of the gas and how much work does it do during the expansion.

My answer:

we know Workdone on = - Workdone by, thus workdone by the gas \begin{align} W &= nRT\ln2\\ \Delta E &= C_v\Delta T + nRT\ln2 \end{align} because the process is isothermal: $\Delta T=0$ thus $$\Delta E=nRT\ln2$$.

However my professor stated in class that \begin{align}\Delta E &= C_v\Delta T \\ &= 0 \end{align}

I'm still confused as to how $\Delta E = 0$ and not $\Delta E = nRT\ln2$

I'd really appreciate it if someone could enlighten me on this.

Thanks!

## 1 Answer

First of all for heat exchange in the process you stated, its not CpdT since the process is not a constant pressure process.

Next, when you talk about heat exchange, its not always necessary that the temperature of the system changes, and this is such an example. Here if you use the formula CpdT for heat, you will see that heat exchange is zero but in reality its not. The isothermal process is such wherein all the heat provided gets utilized in doing work by the gas and hence none of it gets utilized in increasing the temperature.

So to summarize, in your problem, isothermal process tells us that the heat absorbed by the system is utilized completely in doubling the volume i.e in doing work. Hence no temp changes and so internal energy change is zero for the ideal gas.

As a side note, in first law problems involving ideal gases, its easier to track temperature changes and thereby internal energy changes. So usually start thinking about temp change and other things can be sorted out easily.