Problem 22 - University Physics with Modern Physics

Question

The kinetic theory of an ideal gas takes into account not only translational motion of atoms or molecules but also, for diatomic and polyatomic gases, vibration and rotation. Will the temperature increase from a given amount of energy being supplied to a monatomic gas differ from the temperature increase due to the same amount of energy being supplied to a diatomic gas? Explain.

Step-by-Step Solution

Answer

Answer: Yes, the temperature increase is dependent on the type of ideal gas. For the same amount of energy supplied, the temperature increase for a monatomic gas will be twice as large as for a diatomic gas, due to the differences in their degrees of freedom and the ways in which energy is distributed.

Step 1: Understanding the difference between monatomic and diatomic gases

A monatomic gas consists of individual atoms, whereas diatomic gas consists of molecules containing two atoms. Consequently, monatomic gases only have translational motion, while diatomic gases have translational, rotational, and vibrational motion.

Step 2: Calculating degrees of freedom for monatomic and diatomic gases

The degrees of freedom are the number of independent ways that gas particles can move. In an ideal gas, monatomic gases have three translational degrees of freedom, corresponding to the motion in x, y, and z directions. Diatomic gases, on the other hand, have three translational degrees, two rotational degrees, and one vibrational degree of freedom (total of 6).

Step 3: Understanding the molar heat capacities

The molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius. The relationship between the molar heat capacity (C), the number of degrees of freedom (f), and the gas constant (R) is given by: $$ C = \frac{f}{2} R $$ For monatomic gases, f = 3, and for diatomic gases, f = 6.

Step 4: Calculating the temperature increase for monatomic and diatomic gases

We can equate the amount of energy (Q) being supplied to each type of gas with the temperature change (∆T): $$ Q = nC\Delta T $$ Where n is the number of moles. In this case, we are given that the amount of energy supplied to both gases (Q) will be identical. Rearranging for the temperature change (∆T): $$ \Delta T = \frac{Q}{nC} $$ Since we want to compare the temperature increase for both gases, we can create a ratio, such that: $$ \frac{\Delta T_{monatomic}}{\Delta T_{diatomic}} = \frac{\frac{Q}{nC_{monatomic}}}{\frac{Q}{nC_{diatomic}}} $$ The number of moles (n) and the energy supplied (Q) remain the same for both gases. Thus, we can simplify this further by considering only their molar heat capacities' ratios: $$ \frac{\Delta T_{monatomic}}{\Delta T_{diatomic}} = \frac{C_{diatomic}}{C_{monatomic}} $$ Substitute the molar heat capacities formulas in terms of degrees of freedom: $$ \frac{\Delta T_{monatomic}}{\Delta T_{diatomic}} = \frac{\frac{6}{2} R}{\frac{3}{2} R} = 2 $$

Step 5: Conclusion

The ratio of the temperature increase for a monatomic gas to a diatomic gas is 2. This means that the temperature increase for a monatomic gas will be twice as large as for a diatomic gas when they are supplied with the same amount of energy. This is because a monatomic gas only has translational degrees of freedom, whereas diatomic gas has translational, rotational, and vibrational degrees of freedom, resulting in more ways to distribute the energy leading to a smaller temperature increase.