# Pressure variation of gas

1. Jul 20, 2011

### collectedsoul

Work done by a gas = PV

But when we derive specific heat of the gas at constant volume, even though the pressure changes we take work W=0. Or in the case where both pressure and volume are changing and we want to find work done we take the integral of d(PV) where we replace P with nkT/V. Also when we want to find change in heat content $\Delta$Q, we equate it to internal energy $\Delta$E + $\Delta$(PV). If volume is kept constant we eliminate the term $\Delta$(PV). So does that mean that its not possible to add heat to a gas and have it all go to the change in pressure of the gas, i.e., a hypothetical situation where internal energy does not change but pressure does change?

My second question is - is there ever a scenario where dP needs to be taken as an independent variable to find work done or to find some other characteristic of the gas (like Cv)?

Another thing I'm confused about is the case of an isothermal expansion. When the gas expands by a volume dV we have to put in heat to keep it at the same temperature as before. If we do that, is it possible that in the final equilibrium state the pressure might be different from what it was initially? Or is constant pressure guaranteed by the condition of maintaining the same temperature post expansion?

2. Jul 21, 2011

### nasu

The work done by a gas is not PV.
For the special case of a process at constant pressure you can use $$W=p\Delta V$$
In general, you need to integrate $$dW=p dV$$ over the specific process to find the work.

3. Jul 22, 2011

### collectedsoul

Is anything that we are interested in require an integral over small changes in pressure?

4. Jul 23, 2011

### sophiecentaur

Work involves a force moving through a distance. If you don't change the volume then there is no distance moved - so no work. Just in the same way that no work is done as you add more weights onto the top of a table - if it doesn't flex at all.