Q.On a $P$-$V$ diagram, one mole of a perfect gas in a cylindrical container is taken from state 1 (pressure $P_1$, volume $V_1$, temperature $T_1$) to state 2 (pressure $P_2$, volume $V_2$, temperature $T_2$) along a curve for which $P\,V^{1/2}=\text{constant}$. State 1 is at the upper-left (higher pressure, smaller volume) and state 2 at the lower-right (lower pressure, larger volume).

First Law of Thermodynamics

The Intuition: Energy is a Bank Account

Imagine you have a bank account. You can deposit money into it, withdraw money from it, or leave it untouched. The total amount of money in your account changes only when money goes in or comes out. You cannot create money from nothing, and money does not vanish into thin air.

Energy works exactly the same way. In any physical or chemical process, energy is never created and never destroyed. It only moves from one place to another, or changes from one form into another. This is the deepest idea behind the First Law.

Now, in thermodynamics, we focus on a specific "bank account": the internal energy of a system. Internal energy ($U$) is the total energy stored inside a substance — the kinetic energy of its molecules jiggling around, plus the potential energy stored in the bonds between them.

If you want to change how much energy is stored inside a system, you have exactly two ways to do it:

1. Heat ($Q$) — energy that flows because of a temperature difference. Like putting a cold pan on a hot stove.
2. Work ($W$) — energy transferred by a force moving something. Like pushing a piston to compress a gas.

That's it. No third option. Every change in internal energy comes from either heat or work.

The Precise Statement

Where:

• $\Delta U$ = change in internal energy of the system
• $Q$ = heat added to the system (positive if heat flows in)
• $W$ = work done by the system (positive if the system does work on surroundings)

This sign convention is the standard one used in Indian exams (JEE, NEET, etc.). Heat added to the system is positive. Work done by the system is positive.

What This Equation Really Says

Think of it as a balance sheet:

• If you add heat ($Q > 0$), internal energy tends to increase.
• If the system does work ($W > 0$), internal energy tends to decrease (because energy leaves the system to do the work).
• The net change is simply: what came in minus what went out.

If $\Delta U = 0$, the system has returned to its original internal energy — but that does not mean nothing happened. Heat could have come in, and exactly the same amount of energy could have left as work. The energy just passed through.

A Concrete Example

Take a gas trapped in a cylinder with a movable piston. You place the cylinder on a hot plate.

• Heat $Q = +100\ \text{J}$ flows into the gas.
• The gas expands, pushing the piston upward, doing work $W = +40\ \text{J}$ on the surroundings.

What happens to the internal energy?

$\Delta U = 100 - 40 = +60\ \text{J}$ …