One gram mole of oxygen at 27°C and one atmospheric pressure is enclosed in a vessel. (a) Assuming t

Q: Why do we use $v_{\text{rms}}$ for all molecules instead of the actual velocity distribution?

The problem explicitly asks us to assume that all molecules are moving with the root mean square speed for simplicity in calculating the collision rate.

Q: Why is the change in momentum $2mv_{\text{rms}}$?

We assume elastic collisions with the wall. A molecule strikes with momentum $mv_{\text{rms}}$ and rebounds with $-mv_{\text{rms}}$, making the change $2mv_{\text{rms}}$.

Q: Why does the temperature of the gas rise when the vessel stops?

The vessel's sudden stop means its macroscopic kinetic energy is transferred to the gas molecules, increasing their random microscopic kinetic energy, which manifests as a rise in temperature.

Q: What is the value of $C_V$ for oxygen?

Oxygen ($O_2$) is a diatomic gas. At room temperature, it has 5 degrees of freedom (3 translational, 2 rotational), so its molar heat capacity at constant volume is $\frac{5}{2}R$.

Q: How do we calculate the mass of a single oxygen molecule?

The mass of a single molecule is the molar mass divided by Avogadro's number: $m = \frac{32 \times 10^{-3} \text{ kg/mol}}{6.02 \times 10^{23} \text{ mol}^{-1}}$.

Question

One gram mole of oxygen at 27°C and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with $v_{	ext{rms}}$, find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed $v_0$. It is then suddenly stopped. The process results in a rise of the temperature of the gas by 1°C. Calculate the speed $v_0$.

Sigma Percentile · Accepted Answer

The problem of calculating the collision rate of gas molecules and understanding the macroscopic effects of sudden stops is a classic application of the Kinetic Theory of Gases and the First Law of Thermodynamics. Let's break down this fascinating problem into two distinct parts.

### Analyzing the Setup

We are given one mole of oxygen ($O_2$) gas enclosed in a vessel at a temperature of $27^\circ	ext{C}$ (which is $300	ext{ K}$) and a pressure of $1	ext{ atm}$ ($1.01 	imes 10^5	ext{ N/m}^2$).

In the first part, we need to find the number of collisions per second per square meter of the vessel wall, assuming all molecules move with the root mean square speed, $v_{	ext{rms}}$. In the second part, the vessel is moved with a constant speed $v_0$ and suddenly stopped, causing a $1^\circ	ext{C}$ rise in temperature. We need to find $v_0$.

### The Master Equation for Collision Rate

First, let's calculate the root mean square speed of the oxygen molecules. The formula is:

$$v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$$

Substituting the universal gas constant $R = 8.31	ext{ J/(mol K)}$, temperature $T = 300	ext{ K}$, and molar mass $M = 32 	imes 10^{-3}	ext{ kg/mol}$:

$$v_{	ext{rms}} = \sqrt{\frac{3 	imes 8.31 	imes 300}{32 	imes 10^{-3}}} = 483.4	ext{ m/s}$$

Now, let's relate this to pressure. Pressure is defined as the force exerted per unit area. According to the kinetic theory, when a molecule strikes the wall and rebounds elastically, the change in its momentum is $2mv_{	ext{rms}}$.

If $n$ molecules strike one square meter of the wall per second, the total change in momentum per second per unit area (which is the pressure $p_0$) is:

$$p_0 = n 	imes 2mv_{	ext{rms}}$$

### Final Calculation for Part (a)

We can rearrange this equation to solve for the collision rate $n$:

$$n = \frac{p_0}{2mv_{	ext{rms}}}$$

The mass of a single oxygen molecule, $m$, is its molar mass divided by Avogadro's number ($N_A = 6.02 	imes 10^{23}	ext{ mol}^{-1}$):

$$m = \frac{32 	imes 10^{-3}}{6.02 	imes 10^{23}}	ext{ kg}$$

Substituting all the values into our equation for $n$:

$$n = \frac{1.01 	imes 10^5}{2 	imes \left( \frac{32 	imes 10^{-3}}{6.02 	imes 10^{23}} ight) 	imes 483.4}$$

$$n = 1.96 	imes 10^{27}	ext{ s}^{-1}$$

This is an unimaginably large number, highlighting the intense microscopic activity happening constantly within any gas!

### Energy Conservation in a Sudden Stop

Now, let's tackle part (b). The vessel is moving with a speed $v_0$ and suddenly stops. Because the vessel is thermally insulated, no heat escapes. The macroscopic kinetic energy of the gas must be conserved, converting entirely into internal energy.

**Loss in Kinetic Energy = Gain in Internal Energy**

$$\frac{1}{2} m_{	ext{gas}} v_0^2 = n_{	ext{moles}} C_V \Delta T$$

For diatomic oxygen, the molar heat capacity at constant volume is $C_V = \frac{5}{2} R$. We are given $n_{	ext{moles}} = 1$ and $\Delta T = 1^\circ	ext{C}$ (or $1	ext{ K}$).

Rearranging for $v_0$:

$$v_0 = \sqrt{\frac{2 n_{	ext{moles}} C_V \Delta T}{m_{	ext{gas}}}}$$

$$v_0 = \sqrt{\frac{2 	imes 1 	imes \left(\frac{5}{2} 	imes 8.31ight) 	imes 1}{32 	imes 10^{-3}}}$$

$$v_0 = 36	ext{ m/s}$$

This elegant result shows how macroscopic motion directly translates into microscopic thermal energy, a fundamental concept in thermodynamics.

Visualized Solution

Visualizing the Setup

Root Mean Square Speed

Calculating $v_{rms}$

Pressure and Collision Rate

Solving for $n$

Computing Collision Rate

Kinetic Energy to Heat

Calculating $v_{0}$

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation for Collision Rate

Final Calculation for Part (a)

Energy Conservation in a Sudden Stop

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