Calculating Thermodynamic Heat and Work

[Youngster]

Homework Statement

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

Homework Equations

ΔE = q + w
w = -$\int$PdV
H = E + PV

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?

 

[DrClaude]

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

 

[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?

 

[DrClaude]

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?

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

That would mean this is reversible too, yes?

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.

 

[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.

 

[DrClaude]

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).