Chapter 18, Problem 1PE

Molecules of an ideal gas have no intermolecular attractions and, therefore, undergo no change in potential energy on expansion of the gas. If the expansion is also at a constant temperature, there is no change in the kinetic energy, so the isothermal (constant temperature) expansion of an ideal gas has $\Delta E=0$. Suppose such a gas expands at constant temperature from a volume of 1.0 L to 12.0 L against a constant opposing pressure of 14.0 atm. In units of L atm, what are q and w for this change? (Hints: Is the system doing work, or is work done on the system? What must be the sum of q and w in this case?)

Interpretation Introduction

Interpretation:

The values of $q$ and $w$ are to be calculated in units of $\text{Latm}$ for the isothermal expansion of an ideal gas for which the change in volume has been given.

Concept Introduction:

Overall change in the energy of a system is the sum of work done and heat transfer that occurs during the process.

Total energy of a gaseous system is the sum of the potential and kinetic energies possessed by the moecules of the system.

Answer to Problem 1PE

Solution: $-154\text{ L atm}$, $+154\text{ L atm}$

Explanation of Solution

Given information: $\Delta E$ for the isothermal expansion of an ideal gas is equal to $0$. The gas expands at a constant temperature from a volume of $1.0\text{ L}$ to $12.0\text{ L}$ against an opposing pressure of $14.0\text{ atm}$.

Work done during an isothermal expansion can be calculated with the help of the following expression:

$w=-P\Delta V$

Here, w denotes the work done during the process, $P$ denotes the pressure of the gaseous system, and $\Delta V$ denotes the change in volume of the system.

In the given process, the pressure of the gas is $14.0\text{ atm}$ and the volume changes from $1.0\text{ L}$ to $12.0\text{ L}$. On substituting the values of different quantities in the expression, the value of work done in the process can be calculated as:

$\begin{array}{c}w=-(14.0\text{ atm})(12.0\text{ L}-1.0\text{ L})\\ =-154\text{ L atm}\end{array}$

Now, the net change in the energy of a system undergoing a change is given as:

$\Delta E=w+q$

Here, $\Delta E$ represents the net change in energy of the system, $w$ represents the work done, and $q$ represents the heat transferred during the process.

In the given process, $\Delta E=0$. Putting the value of $\Delta E$ and $w$ in the above expression, the value of $q$ is obtained as:

$\begin{array}{l}0=w+q\\ 0=-154+q\\ q=+154\text{ L atm}\end{array}$

Therefore, the values of $q$ and $w$ are calculated.

Conclusion

The values of $q$ and $w$ for the given isothermal expansion are $+154\text{ L atm}$ and $-154\text{ L atm}$, respectively.