First Law of Thermodynamics

Work Done by a Gas

Consider an ideal gas enclosed in a cylinder with a movable, massless and frictionless piston; the gas is the system and the cylinder with its piston is the environment (Fig. 4.6 and Fig. 4.7).

• During expansion the gas molecules that strike the piston lose momentum to it and exert a pressure on it; the piston moves through a finite distance and the gas does positive work on the piston.
• When the piston is pushed in and the volume decreases, the gas molecules striking it gain momentum from it, and the gas does negative work on the piston.

Let the cross-sectional area of the piston be $A$ and the constant pressure exerted by the system on the piston be $p$. The total force exerted by the system on the piston is $F = pA$. If the piston moves through an infinitesimal distance $dx$, the work done by this force is

$$dW = F\,dx = pA\,dx$$

But $A\,dx = dV$, the infinitesimal change in the volume of the cylinder, so

$$dW = p\,dV$$

If the initial volume is $V_i$ and the final volume $V_f$, the total work done in changing the volume is

$$W = \int_{V_i}^{V_f} p\,dV$$

and, when the pressure stays constant during the change,

$$W = p\,(V_f - V_i)$$

Statement of the First Law

If an amount of heat $Q$ is added to a system and the system does no work, its internal energy increases by $\Delta U = Q$. If instead the system does work to increase its volume with no heat added, it loses energy to the surroundings, so $\Delta U = -W$. In practice both happen together, and the total change in internal energy is

$$\Delta U = Q - W$$

This is the mathematical statement of the first law of thermodynamics: the change in the internal energy of a system is the difference between the heat supplied to the system and the work done by the system on its surroundings.

Rearranged,

$$Q = \Delta U + W$$

which is also the first law: when an amount of heat $Q$ is added to a system, its internal energy is increased by an amount $\Delta U$ and the remaining is lost in the form of work done $W$ on the surroundings. Both forms are used and either is acceptable as the statement.

The Three Cases

$Q$ and $W$ can each be positive, negative or zero, and therefore so can $\Delta U$. Figure 4.8 shows the three cases:

• Fig. 4.8 (a): more heat is added to the system than the work done by it, so the internal energy increases, $\Delta U > 0$.
• Fig. 4.8 (b): more work is done by the system than the heat added to it, so the internal energy decreases, $\Delta U < 0$.
• Fig. 4.8 (c): the heat added and the work done are equal, so the internal energy is unchanged, $\Delta U = 0$.

Significance

• The law of conservation of energy studied in Std XI applied to an isolated system, in which there is no exchange of energy at all. The first law is an extension of that law to systems which are not isolated, that is, which can exchange energy as work $W$ or as heat $Q$. The first law is therefore a generalisation of the law of conservation of energy.
• It also makes internal energy measurable. Internal energy defined microscopically from kinetic theory cannot be calculated in practice; $\Delta U = Q - W$ expresses it as the difference of two measurable quantities, the heat supplied to (or released by) the system and the work done by (or on) the system.
• The first law gives the relationship between the heat transfer, the work done, and the change in the internal energy of a system.

Working With Signs in Numericals

The safest routine for the board's numericals is to write down the sign of each quantity before substituting.

• Heat absorbed by the system: $Q$ positive. Heat released by the system: $Q$ negative.
• Work done by the system (expansion): $W$ positive. Work done on the system (compression): $W$ negative.
• Board's worked form when heat is released and work is done on the gas: $\Delta U = |Q| - |W|$. For example, if 125 kJ of heat is released while 104 kJ of work is done on the system, then $Q = -125$ kJ and $W = -104$ kJ, so $\Delta U = -125 - (-104) = -21$ kJ, a decrease of 21 kJ. The board writes exactly this step as $\Delta U = |Q| - |W| = (125 - 104) = 21$ kJ and states the answer as "change in internal energy is 21 kJ", so quote the magnitude and say whether it is an increase or a decrease.
• A gas compressed at constant pressure has $W$ negative. With $W = p\,(V_2 - V_1)$ this fixes $V_2 < V_1$ automatically.

Standard Results

$$dW = p\,dV, \qquad W = \int_{V_i}^{V_f} p\,dV, \qquad W = p\,(V_f - V_i) \ \text{at constant } p$$

$$\Delta U = Q - W \qquad \text{and} \qquad Q = \Delta U + W$$

• First law statement: the change in the internal energy of a system is the difference between the heat supplied to the system and the work done by the system on its surroundings.