Animated Solution for Chemistry - Chemical Thermodynamics: An ideal gas expands in volume from $1imes10^{-3}{\text{ m}}^3$ to $1imes10^{-2}{\text{ m}}^3$ at $300\text{ K}$ against a constant pressure of $1imes10^5{\text{ Nm}}^{-2}$. The work done is

Gas Against the World

Imagine a gas trapped inside a cylinder fitted with a movable piston. The gas is eager to expand, but it has to push against the weight of the atmosphere and the piston itself. This opposing force is what we call the constant external pressure ($p_{\text{ext}}$).

Because the gas is expanding against a fixed, constant pressure rather than a pressure that slowly adjusts to match the gas's internal pressure, this process is classified as an irreversible expansion.

Irreversible Work

In thermodynamics, when a system expands or compresses against a constant external pressure, the work done ($W$) is given by a beautifully simple equation:

Here, $\Delta V$ is the change in volume, calculated as the final volume minus the initial volume ($V_2-V_1$). The negative sign is crucial! It's a sign convention telling us that when the gas expands ($\Delta V$ is positive), the system is doing work on the surroundings, meaning it is losing energy. Hence, the work done by the system is negative.

Watch Your Powers of Ten

Let's plug in the numbers from our problem. We are given: - $p_{\text{ext}}=1\times 10^5{\text{ Nm}}^{-2}$ - $V_1=1\times 10^{-3}{\text{ m}}^3$ - $V_2=1\times 10^{-2}{\text{ m}}^3$

Substituting these into our master equation:

This is where many students make a silly mistake. You cannot directly subtract numbers with different exponents. Let's rewrite $1\times 10^{-2}$ so it has the same exponent as $10^{-3}$.

Now the subtraction is easy:

Now, we just multiply the external pressure by our newly found change in volume:

The unit is Joules because multiplying pressure in ${\text{Nm}}^{-2}$ by volume in ${\text{m}}^3$ gives Newton-meters ($\text{Nm}$), which is exactly equal to Joules. The negative sign confirms our intuition: the gas did $900\text{ J}$ of work to push the piston out.