The work done depends on the type of thermodynamic process. Here are the formulas used in Class 11–12 Chemistry and Physics.

1. Reversible Process

A reversible process is carried out infinitely slowly, so the system remains in equilibrium throughout.

For expansion/compression:$$W = -\int_{V_1}^{V_2} P\,dV$$

If the process is reversible and isothermal (ideal gas):$$W = -nRT\ln\left(\frac{V_2}{V_1}\right)$$

or$$W = -2.303\,nRT\log\left(\frac{V_2}{V_1}\right)$$

---

2. Irreversible Process

If the external pressure is constant:$$W = -P_{\text{ext}}(V_2 – V_1)$$

Special case (free expansion):$$P_{\text{ext}} = 0$$$$W = 0$$

---

3. Isothermal Process (Temperature Constant)

For an ideal gas:$$\Delta U = 0$$

Therefore,$$q = -W$$

Work done (reversible):$$W = -nRT\ln\left(\frac{V_2}{V_1}\right)$$

or$$W = -2.303\,nRT\log\left(\frac{V_2}{V_1}\right)$$

---

4. Adiabatic Process (No Heat Exchange)

$$q = 0$$$$\Delta U = W$$

For a reversible adiabatic process:$$PV^\gamma = \text{constant}$$

and$$W=\frac{P_2V_2-P_1V_1}{\gamma-1}$$

or$$W=nC_V(T_2-T_1)$$

where$$\gamma=\frac{C_P}{C_V}$$

NEET/JEE Perspective

A linear process. It is not automatically reversible or irreversible. If the problem asks for work, calculate the area under the straight line:$$W=\frac{P_1+P_2}{2}(V_2-V_1)=5PV.$$

Quick Revision Table (NCERT)

Important NCERT Facts

• Reversible expansion gives maximum work.
• Irreversible expansion gives less work than reversible expansion.
• Free expansion: $W = 0$W=0.
• Isothermal (ideal gas): $\Delta U = 0$ΔU=0.
• Adiabatic: $q = 0$q=0.