Exercise 3. Calculate internal energy variation by heating the air inside a balloon.

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from $4.00x10^6\;L$ to $4.50x10^6\;L$ by the addition of $1.3x10^8\;J$ of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate internal energy variation for the process. (1 atm = 101.3 J.)


We calculate de work with expresion: \begin{equation} w=-p\Delta V \end{equation} \begin{equation} w=-1\;atm\cdot(4.5x10^6-4x10^6)\;L=-5x10^5\;atm\cdot L \end{equation} We must change the work to joules. \begin{equation} w=-5x10^5\;atmL\cdot\frac{101.3\;J}{1\;atm\cdot L}=-5.1x10^7\;J \end{equation} To calculate $\Delta U$, we use the expresion: \begin{equation} \Delta U=q+w \end{equation} \begin{equation} \Delta U=1.3x10^8-5.1x10^7=8x10^7\;J \end{equation}