Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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 – $dU$.

share|cite|improve this answer
Additionally, if you heat the gas and don't allow it to expand at all, no work is done and all the energy becomes internal energy! – krs013 May 29 '13 at 4:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.