- Imagine a gas in a cylinder (with a piston) where the pressure of
the gas and the external pressure are the same.
- There is no spontaneous change of the gas towards seeking larger or
smaller volume
- Since the internal pressure (\(P\))
and the external pressure (\(P{ext}\))
are the same \[{dw}=-P_{ext}{dV}=-P{dV}\]
Solving Reversible Gas Expansions
\[{dw}=-P_{\text{ext}}{dV}=-P{dV}\]
\[
\begin{align}
w &= -\int {PdV} \\
&= -\int \left ( \frac{nRT}{V} \right ) \\
&= -{nRT} \int_{V_1}^{V_2} \frac{dV}{V} =
-{nRT}\ln\frac{V_2}{V_1} \\
\end{align}
\]
Example 3.3
What is the work done by \(1.00{mol}\) of an ideal gas expanding
reversibly from a volume of \(22.4{L}\)
to a volume of \(44.8{L}\) at a
constant temperature of \(273K\)?
Note: Isothermal Processes of an Ideal Gas
- Isothermal means the process occurs at a constant temperature
- Ideal gases, due to its assumptions, can only store energy as
kinetic energy
- Changing kinetic energy means changing temperature
- So, no change in temperature means no change in \(U\)
- \(\Delta U = 0\therefore q = -w\)
for isothermal processes involving ideal gases