# Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

### 3. Answer in brief.

vii. Derive the expression for PV work

- Imagine a specific quantity of gas held at a constant pressure \(P\) inside a cylinder equipped with a frictionless, rigid piston of area \(A\). The gas initially occupies a volume \(V_1\) at temperature \(T\), as illustrated in the adjacent diagram.
- During expansion, the force exerted by the gas is equal to the area of the piston multiplied by the pressure at which the gas acts against the piston. This pressure has the same magnitude as the external atmospheric pressure but opposite in sign, denoted as \(-P_{\text{ext}}\). Hence,
- When the piston moves a distance \(d\), the work performed is the product of the force and the distance traveled:
- The product of the piston's area and the distance it moves represents the change in volume (\(\Delta V\)) of the system:

\[f = -P_{\text{ext}} \cdot A \quad \text{(1)}\]

Here, \(P_{\text{ext}}\) represents the external atmospheric pressure.

\[W = f \cdot d \quad \text{(2)}\]

By substituting equation (1) into equation (2), we obtain:

\[W = -P_{\text{ext}} \cdot A \cdot d \quad \text{(3)}\]

\[\Delta V = A \cdot d \quad \text{(4)}\]

Combining equations (3) and (4), we get:

\[W = -P_{\text{ext}} \cdot \Delta V\]

\[W = -P_{\text{ext}} \cdot (V_2 - V_1)\]

Here, \(V_2\) denotes the final volume of the gas.

Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution