# Calculating Thermodynamic Heat and Work

1. Aug 18, 2013

### Youngster

1. The problem statement, all variables and given/known data

One mole of ideal gas expands isothermally at 20°C against a constant pressure of 1 atm from 20 to 30L. (1 atm = 1.013 x 105Pa; 1 m3 = 1000L

Find q, w, ΔE, and ΔH in Joules

2. Relevant equations

ΔE = q + w
w = -$\int$PdV
H = E + PV

3. The attempt at a solution

I solved for work using the second equation listed above. Since pressure is constant, I removed it from the integral and simply integrated the volume from 20L to 30L. This resulted in -10 atm x L.

Converting that to Joules using the conversion factors given, I obtained -1013 J as the thermodynamic work.

But then how do I obtain the thermodynamic heat?

2. Aug 19, 2013

### Staff: Mentor

You need to figure out the change in the internal energy of the gas, $\Delta E$, considering it is ideal.

3. Aug 20, 2013

### Youngster

Okay, so what I'm getting is that since it's an isothermal process, the change in internal energy is 0.

That would mean q is just the reverse of w. Does this sound right? That would mean this is reversible too, yes?

4. Aug 20, 2013

### Staff: Mentor

Yes. Just to be clear, the energy lost by the gas as it does work is compensated by heat coming in from the environment.

I guess that it is reasonable to assume that this is the result of a slow (quasistatic) process, for the temperature to be maintained constant during the entire process, and therefore it is reversible.

5. Aug 20, 2013

### Youngster

Ah yes, thank you.

I have one more item I'd like to clarify.

I have two equation for thermodynamic work listed in my notes.

1. w = -∫PdV
2. w = -nRT ln$\frac{V_{2}}{V_{2}}$

I used the first equation to obtain -1013J, but the second equation provides a different value of -987.7J. This is a pretty significant difference, so I'm curious what the differences between the equations are, if any.

6. Aug 20, 2013

### Staff: Mentor

I didn't pay enough attention to the original statement. The situation described is simply impossible: an ideal gas cannot expand when both temperature and pressure are held constant. Obviously, since $PV = nR T$, $V$ can't change if $P$, $T$, and $n$ are all kept constant.

Otherwise, you get equation 2 from equation 1 by replacing $P$ by $n R T/V$, and then integrating over $V$ for $T$ constant. Equation 2 is therefore useful for isothermal processes (for an ideal gas only).