3.3 First Law of Thermodynamics

Changes of State and the First Law
During a thermodynamic process, a system moves from state A to state B, it is supplied with 400 J of heat and does 100 J of work.

(a) For this transition, what is the system’s change in internal energy?
(b) If the system then moves from state B back to state A, what is its change in internal energy?
(c) If in moving from A to B along a different path, ${W^{\prime }}_{AB}=400J$ of work is done on the system, how much heat does it absorb?

Strategy
The first law of thermodynamics relates the internal energy change, work done by the system, and the heat transferred to the system in a simple equation. The internal energy is a function of state and is therefore fixed at any given point regardless of how the system reaches the state.

Solution

a. From the first law, the change in the system’s internal energy is

$\mathrm{\Delta }{E}_{\text{int}AB}=Q_{AB}-W_{AB}=400\text{J}-100\text{J}=300\text{J}.$

b. Consider a closed path that passes through the states A and B. Internal energy is a state function, so $\mathrm{\Delta }{E}_{\text{int}}$ is zero for a closed path. Thus

$\mathrm{\Delta }{E}_{\text{int}}=\mathrm{\Delta }{E}_{\text{int}AB}+\mathrm{\Delta }{E}_{\text{int}BA}=0,$

and

$\mathrm{\Delta }{E}_{\text{int}AB}=-{\mathrm{\Delta }}_{\text{int}BA}.$

This yields

$\mathrm{\Delta }{E}_{\text{int}BA}=-300\text{J}.$

c. The change in internal energy is the same for any path, so

$\begin{array}{rcl}\mathrm{\Delta }{E}_{intAB}& =& \mathrm{\Delta }{E^{\prime }}_{intAB}={Q^{\prime }}_{AB}-{W^{\prime }}_{AB};\\ 300J& =& {Q^{\prime }}_{AB}-(-400J),\end{array}$

and the heat exchanged is

$Q{\prime }_{AB}=-100\text{J}.$

The negative sign indicates that the system loses heat in this transition.

Significance
When a closed cycle is considered for the first law of thermodynamics, the change in internal energy around the whole path is equal to zero. If friction were to play a role in this example, less work would result from this heat added. Example 3.3 takes into consideration what happens if friction plays a role.

Polishing a Fitting
A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth across the fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting.

(a) What is the total work done on the fitting by the machinist?
(b) What is the increase in the internal energy of the fitting? Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between the fitting and its environment.
(c) What is the increase in the temperature of the fitting?

Strategy
The machinist’s force over a distance that can be calculated from the speed and time given is the work done on the system. The work, in turn, increases the internal energy of the system. This energy can be interpreted as the heat that raises the temperature of the system via its heat capacity. Be careful with the sign of each quantity.

Solution

a. The power created by a force on an object or the rate at which the machinist does frictional work on the fitting is $\overrightarrow{F}\cdot \overrightarrow{v}=-Fv$. Thus, in an elapsed time $\mathrm{\Delta }t$ (2.0 min), the work done on the fitting is

$\begin{array}{rcl}W& =& -Fv\mathrm{\Delta }t=-(20N)(0.1m/s)(1.2\times {10}^{2}s)\\ & =& -2.4\times {10}^{3}J.\end{array}$

b. By assumption, no heat is exchanged between the fitting and its environment, so the first law gives for the change in the internal energy of the fitting:

$\mathrm{\Delta }{E}_{\text{int}}=-W=2.4\times {10}^3\text{J}.$

c. Since $\mathrm{\Delta }{E}_{\text{int}}$ is path independent, the effect of the $2.4\times {10}^3\text{J}$ of work is the same as if it were supplied at atmospheric pressure by a transfer of heat. Thus,

$2.4\times {10}^3\text{J}=mc\mathrm{\Delta }T=(0.50\text{kg})(3.9\times 102\text{J/kg}{\cdot }^{\circ }\text{C})\mathrm{\Delta }T,$

and the increase in the temperature of the fitting is

$\mathrm{\Delta }T={12}^{\circ }\text{C}$

where we have used the value for the specific heat of copper, $c=3.9\times {10}^2{\text{J/kg}}^{\circ }\text{C}$.

Significance
If heat were released, the change in internal energy would be less and cause less of a temperature change than what was calculated in the problem.

An Ideal Gas Making Transitions between Two States
Consider the quasi-static expansions of an ideal gas between the equilibrium states A and C of Figure 3.6. If 515 J of heat are added to the gas as it traverses the path ABC, how much heat is required for the transition along ADC? Assume that $p_1=2.10\times {10}^5{\text{N/m}}^2$, $p_2=1.05\times {10}^5{\text{N/m}}^2$, $V_1=2.25\times {10}^{-3}{m}^3$, and $V_2=4.50\times {10}^{-3}{m}^3$.

Strategy
The difference in work done between process ABC and process ADC is the area enclosed by ABCD. Because the change of the internal energy (a function of state) is the same for both processes, the difference in work is thus the same as the difference in heat transferred to the system.

Solution
For path ABC, the heat added is $Q_{ABC}=515\text{J}$ and the work done by the gas is the area under the path on the pV diagram, which is

$W_{ABC}=p_1(V_2-V_1)=473\text{J}.$

Along ADC, the work done by the gas is again the area under the path:

$W_{ADC}=p_2(V_2-V_1)=236\text{J}.$

Then using the strategy we just described, we have

$Q_{ADC}-Q_{ABC}=W_{ADC}-W_{ABC},$

which leads to

$Q_{ADC}=Q_{ABC}+W_{ADC}-W_{ABC}=(515+236-473)\text{J}=278\text{J}.$

Significance
The work calculations in this problem are made simple since no work is done along AD and BC and along AB and DC; the pressure is constant over the volume change, so the work done is simply $p\mathrm{\Delta }V$. An isothermal line could also have been used, as we have derived the work for an isothermal process as $W=nRT\text{ln}\frac{V_2}{V_1}$.

Isothermal Expansion of an Ideal Gas
Heat is added to 1 mol of an ideal monatomic gas confined to a cylinder with a movable piston at one end. The gas expands quasi-statically at a constant temperature of 300 K until its volume increases from V to 3V.

(a) What is the change in internal energy of the gas?
(b) How much work does the gas do?
(c) How much heat is added to the gas?

Strategy

(a) Because the system is an ideal gas, the internal energy only changes when the temperature changes.
(b) The heat added to the system is therefore purely used to do work that has been calculated in Work, Heat, and Internal Energy.
(c) Lastly, the first law of thermodynamics can be used to calculate the heat added to the gas.

Solution

a. We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only of temperature. Since $\mathrm{\Delta }T=0$, for this process, $\mathrm{\Delta }{E}_{\text{int}}=0$.

b. The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and was found to be

$\begin{array}{rl}W& =nRT\ln \frac{V_2}{V_1}=nRT\ln \frac{3V}{V}\\ & =(1.00\text{mol})(8.314\text{J/K}\cdot \text{mol})(300\text{K})(\ln 3)=2.74\times {10}^3\text{J}.\end{array}$

c. With the results of parts (a) and (b), we can use the first law to determine the heat added:

$\mathrm{\Delta }{E}_{\text{int}}=Q-W=0,$

which leads to

$Q=W=2.74\times {10}^3\text{J}.$

Significance
An isothermal process has no change in the internal energy. Based on that, the first law of thermodynamics reduces to $Q=W$.

Vaporizing Water
When 1.00 g of water at 100°C changes from the liquid to the gas phase at atmospheric pressure, its change in volume is $1.67\times {10}^{-3}{\text{m}}^3$.

(a) How much heat must be added to vaporize the water?
(b) How much work is done by the water against the atmosphere in its expansion?
(c) What is the change in the internal energy of the water?

Strategy
We can first figure out how much heat is needed from the latent heat of vaporization of the water. From the volume change, we can calculate the work done from $W=p\mathrm{\Delta }V$ because the pressure is constant. Then, the first law of thermodynamics provides us with the change in the internal energy.

Solution

a. With L_v representing the latent heat of vaporization, the heat required to vaporize the water is

$Q=m{L}_v=(1.00\text{g})(2.26\times {10}^3\text{J/g})=2.26\times {10}^3\text{J}.$

b. Since the pressure on the system is constant at $1.00\text{atm}=1.01\times {10}^5{\text{N/m}}^2$, the work done by the water as it is vaporized is

$W=p\mathrm{\Delta }V=(1.01\times {10}^5{\text{N/m}}^2)(1.67\times {10}^{-3}{m}^3)=169\text{J}.$

c. From the first law, the thermal energy of the water during its vaporization changes by

$\mathrm{\Delta }{E}_{\text{int}}=Q-W=2.26\times {10}^3\text{J}-169\text{J}=2.09\times {10}^3\text{J}.$

Significance
We note that in part (c), we see a change in internal energy, yet there is no change in temperature. Ideal gases that are not undergoing phase changes have the internal energy proportional to temperature. Internal energy in general is the sum of all energy in the system.