Consider that an ideal gas ($n$ moles) is expanding in a process given by $P = f(V)$, which passes through a point ($V_0$, $P_0$). Show that the gas is absorbing heat at ($P_0$, $V_0$) if the slope of the curve $P = f(V)$ is larger than the slope of the adiabat passing through ($P_0$, $V_0$).

> [!TLDR] > The condition for heat absorption is derived from the First Law: $dQ = dU + PdV$. By comparing the slope of the given process with the adiabatic slope at the same point, we show that $dQ > 0$ exactly when $f'(V_0) > -\gamma P_0/V_0$. The core idea is simple: heat absorbed by a gas equals the change in internal energy plus the work done. For an ideal gas, internal energy depends only on temperature, so we can connect the slope of the $P$-$V$ curve to the temperature change and hence to heat flow. 1. **Start with the First Law** For an infinitesimal step, $dQ = dU + PdV$. For an ideal gas, $dU = nC_v dT$. Using the ideal gas law $PV = nRT$, we can write $dT$ in terms of $P$ and $V$: $$dT = \frac{1}{nR}(PdV + VdP)$$ Substituting: $$dQ = nC_v \cdot \frac{1}{nR}(PdV + VdP) + PdV = \frac{C_v}{R}(PdV + VdP) + PdV$$ 2. **Simplify using $C_p - C_v = R$** Since $\frac{C_v}{R} = \frac{C_v}{C_p - C_v} = \frac{1}{\gamma - 1}$, where $\gamma = C_p/C_v$, we get: $$dQ = \frac{1}{\gamma - 1}(PdV + VdP) + PdV = \frac{1}{\gamma - 1}VdP + \frac{\gamma}{\gamma - 1}PdV$$ 3. **Express in terms of slope** At the point $(P_0, V_0)$, the slope of the given process is $f'(V_0) = \left(\frac{dP}{dV}\right)_{\text{process}}$. So $dP = f'(V_0) dV$. Substituting: $$dQ = \left[ \frac{1}{\gamma - 1} V_0 f'(V_0) + \frac{\gamma}{\gamma - 1} P_0 \right] dV$$ 4. **Condition for heat absorption** Heat is absorbed when $dQ > 0$. Assuming $dV > 0$ (expansion), this requires: $$\frac{1}{\gamma - 1} V_0 f'(V_0) + \frac{\gamma}{\gamma - 1} P_0 > 0$$ Multiply through by $(\gamma - 1) > 0$: $$V_0 f'(V_0) + \gamma P_0 > 0 \quad \Rightarrow \quad f'(V_0) > -\frac{\gamma P_0}{V_0}$$ 5. **Interpret the right-hand side** For an adiabatic process on an ideal gas, $PV^\gamma = \text{constant}$. Differentiating: $$V^\gamma dP + \gamma P V^{\gamma-1} dV = 0 \quad \Rightarrow \quad \left(\frac{dP}{dV}\right)_{\text{adiabat}} = -\frac{\gamma P}{V}$$

Consider that an ideal gas (n moles) is expanding in a process given by P = f(V), which passes through a point

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Concept understanding — First Law of Thermodynamics

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:

Heat ( $Q$ ) — energy that flows because of a temperature difference. Like putting a cold pan on a hot stove.
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

\Delta U = Q - W

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.

Watch out

Some textbooks use $Q = \Delta U + W$ or $\Delta U = Q + W$ with a different sign for work. Always check which convention your exam follows. The one above ( $\Delta U = Q - W$ ) is the most common in Indian syllabi.

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.

Note

| Process | $Q$ | $W$ | $\Delta U$ |

|---------|-----|-----|------------|

| Gas heated at constant volume | + | 0 | Positive |

| Gas cooled at constant volume | – | 0 | Negative |

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?

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

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The condition for heat absorption is derived from the First Law: $dQ = dU + PdV$ . By comparing the slope of the given process with the adiabatic slope at the same point, we show that $dQ > 0$ exactly when $f'(V_0) > -\gamma P_0/V_0$ .

The core idea is simple: heat absorbed by a gas equals the change in internal energy plus the work done. For an ideal gas, internal energy depends only on temperature, so we can connect the slope of the $P$ - $V$ curve to the temperature change and hence to heat flow.

Start with the First Law For an infinitesimal step, $dQ = dU + PdV$ . For an ideal gas, $dU = nC_v dT$ . Using the ideal gas law $PV = nRT$ , we can write $dT$ in terms of $P$ and $V$ :

dT = \frac{1}{nR}(PdV + VdP)

Substituting:

dQ = nC_v \cdot \frac{1}{nR}(PdV + VdP) + PdV = \frac{C_v}{R}(PdV + VdP) + PdV

Simplify using $C_p - C_v = R$ Since $\frac{C_v}{R} = \frac{C_v}{C_p - C_v} = \frac{1}{\gamma - 1}$ , where $\gamma = C_p/C_v$ , we get:

dQ = \frac{1}{\gamma - 1}(PdV + VdP) + PdV = \frac{1}{\gamma - 1}VdP + \frac{\gamma}{\gamma - 1}PdV

Express in terms of slope At the point $(P_0, V_0)$ , the slope of the given process is $f'(V_0) = \left(\frac{dP}{dV}\right)_{\text{process}}$ . So $dP = f'(V_0) dV$ . Substituting:

dQ = \left[ \frac{1}{\gamma - 1} V_0 f'(V_0) + \frac{\gamma}{\gamma - 1} P_0 \right] dV

Condition for heat absorption Heat is absorbed when $dQ > 0$ . Assuming $dV > 0$ (expansion), this requires:

\frac{1}{\gamma - 1} V_0 f'(V_0) + \frac{\gamma}{\gamma - 1} P_0 > 0

Multiply through by $(\gamma - 1) > 0$ :

V_0 f'(V_0) + \gamma P_0 > 0 \quad \Rightarrow \quad f'(V_0) > -\frac{\gamma P_0}{V_0}

Interpret the right-hand side For an adiabatic process on an ideal gas, $PV^\gamma = \text{constant}$ . Differentiating:

V^\gamma dP + \gamma P V^{\gamma-1} dV = 0 \quad \Rightarrow \quad \left(\frac{dP}{dV}\right)_{\text{adiabat}} = -\frac{\gamma P}{V}

At $(P_0, V_0)$ , the adiabatic slope is exactly $-\gamma P_0 / V_0$ . …

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