Chapter 5, Problem 5.78P

(a)

Interpretation Introduction

Interpretation:

The curve that represents the behavior of gas at low temperature is to be identified.

Concept introduction:

The expression to calculate the most probable speed is as follows:

  $u_{\text{mps}}=\sqrt{\frac{2RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the root-mean-square speed is as follows:

  $u_{rms}=\sqrt{\frac{3RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the average kinetic energy of gases is as follows:

  $\text{Average kinectic energy of gases}=\left(\frac{3}{2}\right)\left(RT\right)$

Here, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the kinetic energy is as follows:

  $E_{\text{k}}=\left(\frac{1}{2}\right)\left(\text{mass}\right){\left(\text{speed}\right)}^2$

(b)

Interpretation Introduction

Interpretation:

The curve that represents the gas that has higher kinetic energy is to be identified.

Concept introduction:

The expression to calculate the average kinetic energy of gases is as follows:

  $\text{Average kinectic energy of gases}=\left(\frac{3}{2}\right)\left(RT\right)$

Here, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the kinetic energy is as follows:

  $E_{\text{k}}=\left(\frac{1}{2}\right)\left(\text{mass}\right){\left(\text{speed}\right)}^2$

The kinetic energy is directly proportional to the temperature.

(c)

Interpretation Introduction

Interpretation:

The curve consistent with a higher diffusion rate is to be identified.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole. Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  $\frac{\text{Rate of A}}{\text{Rate of B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}$

Here,

$M_{\text{B}}$ is the molar mass of gas $\text{B}$

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.