Adding heat to a system doesn't equal the work done by gas?

Question

So, I answered a physics question for a class that goes as follows:

A gas in a cylinder is kept at a constant pressure of $250000\: \mathrm{Pa}$ while $300\: \mathrm{kJ}$ of heat are added to it, causing the has to expand from $0.9\: \mathrm{m^3}$ to $1.5\: \mathrm{m^3}$. What is the work done by gas?

I knew the answer was to use $W = P \Delta V$, which worked, giving an answer of $2.1 \times 10^5\: \mathrm{J}$.

However, conceptually I am having a hard time understanding why the $300\: \mathrm{kJ}$ of heat being added to the system doesn't just equal the amount of work done. Aren't why just converting the heat energy into work, with all forces conserved?

Answer 1

Because there's another term in the equation called the first law of thermodynamics, namely $dU$, the change of the internal energy! $$ dU = \delta Q - \delta W$$ When one heats an object, the most obvious consequence is that this something gets warmer. It doesn't have to expand to do any work, it just heats up. In a general case, the heat $\delta Q$ is divided to the work done $\delta W$ and the change of the internal energy – dependent largely on temperature – $dU$.