2.4: PASS Ideal Gases- use the Kinetic Molecular Theory to explain impact on a gas as V and T increased (2.E.45)

Exercise 2.E.45

A 1 L sample of CO initially at 0^oC and 1atm is heated to 546 K, and its volume is increased to 2 L.

1. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
2. What is the effect on the average kinetic energy of the molecules?
3. What is the effect on the root mean square speed of the molecules?

Answer

1. The number of collisions per unit area of the container wall is constant.
2. The average kinetic energy doubles.
3. The root mean square speed increases \(\sqrt{2}\) times its initial value; U_rms is proportional to KE_avg.

Strategy Map

Step	Hint
1. Recall the relationships between temperature, pressure, and volume.	As Temperature increases, Volume increases. As Temperature increases , Pressure increases. As Volume increases , Pressure decreases.
2. Recall the relationship between temperature and kinetic energy.	As Temperature increases , the kinetic energy of gas molecules increases. See LibreText section 2.5 The Kinetic-Molecular Theory
3. Recall the relationship to determine average kinetics energy and what change would cause it to increase or decrease.	\(K E_{\text {avg }}=\frac{3}{2} R T\)
4. Recall what root mean square speed is and what causes it to increase or decrease.	\(U_{r m s}=\sqrt{\frac{3 R T}{M(x)}}\)
5. Look at the question information and determine which parameters change and how they are related.	V1 = 1 L; V2 = 2 L T1 = 0^oC = 273 K; T2 = 546 K n is constant What happens to P?

Solution

\(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2}\)

\(\begin{aligned}
& \mathrm{T}_2=2 \mathrm{~T}_1 \\
& \mathrm{~V}_2=2 \mathrm{~V}_2
\end{aligned}\)

Substitute in to the equation for changing conditions:

\(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 2 \mathrm{~V}_1}{2 \mathrm{~T}_1}\)

Cancel terms that are the same

\(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 2 \mathrm{~V}_1}{2 \mathrm{~T}_1}\)

Left with P₁ = P₂, no change in pressure.

answer 1. Since pressure does not change, the number of collisions per unit area of the container wall is constant.

Looking at the impact of increased temperature on kinetic energy, as T doubles kinetic energy doubles.

\(K E_{\text {avg }}=\frac{3}{2} R T\)

answer 2. Since T₂ is 2 times T₁, the average kinetic energy doubles.

Looking at the impact of increased temperature on root mean square speed, as T doubles root mean square speed increase by \(\sqrt{2}\) times.

\(U_{r m s}=\sqrt{\frac{3 R T}{M(x)}}\)

\(U_{r m s}=\sqrt{\frac{3 R T_1}{M(x)}} \text { or } U_{r m s}=\sqrt{\frac{3 R\left(2 T_1\right)}{M(x)}}\)

answer 3. The root mean square speed increases since the temperature has increased, it increases \(\sqrt{2}\) times its initial value; U_rms is proportional to KE_avg.

Guided Solution

Download Guided Solution as a pdf

Guided Solution	Hint
This is a theory type problem that tests your knowledge on the kinetic molecular theory. The temperature and volume of a gas is increased, you must use your knowledge on the relationships of this chapter to predict the kinetics of the gas.	See LibreText section 2.5 The Kinetic-Molecular Theory
A 1 L sample of CO initially at 0^oC and 1atm is heated to 546 K, and its volume is increased to 2 L. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall? What is the effect on the average kinetic energy of the molecules? What is the effect on the root mean square speed of the molecules?	As the temperature of the gas was increased from 273 K to 546 K, the volume of the gas increased from 1L to 2L.
Recall your relationships	As temperature increases, pressure increases. As temperature increases, volume increases. As temperature increases, kinetic energy of gas molecules increases. As volume increases, pressure decreases.
Recall the equations that demonstrate these relationships	PV = nRT \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2}\) \(K E_{\text {avg }}=\frac{3}{2} R T\)
What is “root mean square speed” and what equation is used to calculate it?	The root mean square speed of a particle, Urms, is defined as the square root of the average of the squares of the velocities with n = the number of particles. \(U_{r m s}=\sqrt{\frac{3 R T}{M(x)}}\)
Complete Solution: \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2}\) \(\begin{aligned} & \mathrm{T}_2=2 \mathrm{~T}_1 \\ & \mathrm{~V}_2=2 \mathrm{~V}_2 \end{aligned}\) Substitute in to the equation for changing conditions: \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 2 \mathrm{~V}_1}{2 \mathrm{~T}_1}\) Cancel terms that are the same: \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 2 \mathrm{~V}_1}{2 \mathrm{~T}_1}\) Left with P1 = P2, no change in pressure. The pressure is the number of collisions per unit of area, if the pressure is constant, so are the number of collisions. answer 1. Since pressure does not change, the number of collisions per unit area of the container wall is constant. \(K E_{\text {avg }}=\frac{3}{2} R T\) As the temperature increases, so does the average kinetic energy. answer 2. Since T2 is 2 times T1, the average kinetic energy doubles. \(U_{r m s}=\sqrt{\frac{3 R T}{M(x)}}\) \(U_{r m s}=\sqrt{\frac{3 R T_1}{M(x)}} \text { or } U_{r m s}=\sqrt{\frac{3 R\left(2 T_1\right)}{M(x)}}\) If the temperature increases, and the number of particles stays constant, the speed will increase. answer 3. The root mean square speed increases since the temperature has increased, it increases \(\sqrt{2}\) times its initial value; Urms is proportional to KEavg.
Check your work! Make sure you look at the impact of increasing temperature and increasing volume on the parameters you are asked about in the question. To evaluate impact on collisions you must first think about impact on pressure using the appropriate form of the general gas law for changing conditions. Since T increases we expect KEavg and Urms to increase. Why does this answer make chemical sense? If one factor causes an increase in pressure while the other causes a decrease in the pressure, they will counteract one another. In this case each increased by a factor of 2 which cancels. So the pressure will stay constant. If factors cause the speed to increase, and nothing impacts the speed to decrease, the particles will move faster. This can be seen by the equations below. \(K E_{\text {avg }}=\frac{3}{2} R T\) This shows that kinetic energy and temperature are proportional. \(U_{r m s}=\sqrt{\frac{3 R T}{M(x)}}\) This shows that the root mean square speed is dependent on temperature.	See LibreText section 2.5 The Kinetic-Molecular Theory

(question source from page titled 9.E:Gases (Exercises) https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_(OpenSTAX)/09%3A_Gases/9.E%3A_Gases_(Exercises), shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source https://openstax.org/books/chemistry/pages/9-exercises, Access for free at https://openstax.org/books/chemistry/pages/1-introduction)