We have seen that the internal energy \(U\) of a system can change through two modes of energy transfer: heat and work. Let
\(\Delta Q=\) Heat supplied to the system by the surroundings
\(\Delta W=\) Work done by the system on the surroundings
\(\Delta U=\) Change in internal energy of the system The general principle of conservation of energy then implies that
\(\Delta Q=\Delta U+\Delta W \dots(12.1)\)
1.e. the energy \((\Delta Q)\) supplied to the system goes in part to increase the internal energy of the system \((\Delta U)\) and the rest in work on the environment \((\Delta W)\). Equation (12.1) is known as the First Law of Thermodynamics. It is simply the general law of conservation of energy applied to any system in which the energy transfer from or to the surroundings is taken into account. Let us put Eq. (12.1) in the alternative form
\(
\Delta Q-\Delta W=\Delta U \dots(12.2)
\)
Now, the system may go from an initial state to the final state in a number of ways. For example, to change the state of a gas from \(\left(P_1, V_1\right)\) to \(\left(P_2, V_2\right)\), we can first change the volume of the gas from \(V_1\) to \(V_2\), keeping its pressure constant i.e. we can first go the state \(\left(P_1, V_2\right)\) and then change the pressure of the gas from \(P_1\) to \(P_2\), keeping the volume constant, to take the gas to \(\left(P_2, V_2\right)\). Alternatively, we can first keep the volume constant and then keep the pressure constant. Since \(U\) is a state variable, \(\Delta U\) depends only on the initial and final states and not on the path taken by the gas to go from one to the other. However, \(\Delta Q\) and \(\Delta W\) will, in general, depend on the path taken to go from the initial to final states.

From the First Law of Thermodynamics, Eq. (12.2), it is clear that the combination \(\Delta Q-\Delta W\), is, however, path independent. This shows that if a system is taken through a process in which \(\Delta U=0\) (for example, isothermal expansion of an ideal gas),
\(
\Delta Q=\Delta W
\)
1.e., heat supplied to the system is used up entirely by the system in doing work on the environment.

Example 1: A gas is contained in a vessel fitted with a movable piston. The container is placed on a hot stove. A total of \(100 \mathrm{cal}\) of heat is given to the gas and the gas does \(40 \mathrm{~J}\) of work in the expansion resulting from heating. Calculate the increase in internal energy in the process.

Solution:

Heat given to the gas is \(\Delta Q=100 \mathrm{cal}=418 \mathrm{~J}\).
Work done by the gas is \(\Delta W=40 \mathrm{~J}\).
The increase in internal energy is
\(
\begin{aligned}
\Delta U & =\Delta Q-\Delta W \\
& =418 \mathrm{~J}-40 \mathrm{~J}=378 \mathrm{~J}
\end{aligned}
\)

Work Done By A Gas

Consider a gas contained in a cylinder of the cross-sectional area \(A\) fitted with a movable piston. Let the pressure of the gas be \(P\). The force exerted by the gas on the piston is \(P A\) in the outward direction. Suppose the gas expands a little and the piston is pushed out by a small distance \(\Delta x\). The work done by the gas on the piston is
\(
\Delta W=(P A)(\Delta x)=P \Delta V,
\)

where \(\Delta V=A \Delta x\) is the change in the volume of the gas. For a finite change of volume from \(V_1\) to \(V_2\), the pressure may not be constant. We can divide the whole process of expansion in small steps and add the work done in each step. Thus, the total work done by the gas in the process is
\(
W=\int_{V_1}^{V_2} p d V
\)
If we show the process in a \(P-V\) diagram, the work done is equal to the area bounded by the \(P-V\) curve, the \(V\)-axis and the ordinates \(V=V_1\) and \(V=V_2\).

Therefore, we can write If the system is a gas in a cylinder with a movable piston, the gas in moving the piston does work. Since force is pressure times area, and area times displacement is volume, work done by the system against a constant pressure \(P\) is
\(
\Delta W=P \Delta V
\)
where \(\Delta V\) is the change in volume of the gas. Thus, for this case, Eq. (12.1) gives
\(
\Delta Q=\Delta U+P \Delta V \dots(12.3)
\)

Example 2: Calculate the work done by a gas as it is taken from the state \(a\) to \(b, b\) to \(c\), and \(c\) to \(a\) as shown in the figure below.

Solution:

The work done by the gas in the process \(a\) to \(b\) is the area of abde. This is
\(
\begin{aligned}
W_{a b} & =(120 \mathrm{kPa})(250 \mathrm{cc}) \\
& =120 \times 10^3 \times 250 \times 10^{-6} \mathrm{~J}=30 \mathrm{~J} .
\end{aligned}
\)
In the process \(b\) to \(c\) the volume remains constant and the work done is zero.

In the process \(c\) to \(a\) the gas is compressed. The volume is decreased and the work done by the gas is negative. The magnitude is equal to the area of caed. This area is \(c a b+b a e d\)
\(
\begin{aligned}
& =\frac{1}{2}(80 \mathrm{kPa})(250 \mathrm{cc})+30 \mathrm{~J} \\
& =10 \mathrm{~J}+30 \mathrm{~J}=40 \mathrm{~J} .
\end{aligned}
\)
Thus, the work done in the process \(c\) to \(a\) is \(-40 \mathrm{~J}\).