Chapter 11, Problem 83UTC

(a)

Interpretation Introduction

Interpretation: This is to determine whether the given statement for the pressure of helium and neon gases present in two separate flasks under the given conditions is true of false.

Concept introduction:

Pressure, volume, temperature and number of moles of a gas are related through a thermodynamic equation of state called the ideal gas equation.

Answer to Problem 83UTC

The given statement is true.

Explanation of Solution

As per the ideal gas equation:

  $PV=nRT$   ....... (1)

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

Here, number of moles is ratio of mass and molar mass thus, above equation can be rewritten as follows:

  $PV=\frac{m}{M}RT$   ........ (2)

According to equation (2), the pressure P of a gas is inversely proportional to its molar mass. The molar mass for $\text{He }=\text{ 4 g}/\text{mol}$ and that of $\text{Ne }=\text{ 2}0\text{ g}/\text{mol}$ . Therefore, $\text{He}$ will have a higher pressure than

  $\text{Ne}$ .

Hence, the given statement is true.

(b)

Interpretation Introduction

Interpretation: This is to determine whether the given statement for thedenisty of helium and neon gases present in two separate flasks under the given conditions is true of false.

Concept introduction:

Density can be calculated from mass and volume as follows:

  $\text{Density (}\rho \text{) = }\frac{\text{mass (m)}}{\text{Volume (V)}}$

Answer to Problem 83UTC

The given statement is true.

Explanation of Solution

The density is given as:

  $\text{Density (}\rho \text{) = }\frac{\text{mass (m)}}{\text{Volume (V)}}$

Since, the mass and volume of both the gases are same. Therefore, ratio of mass and volume that is the density of both the gases are same.

Under the given conditions, $\text{He}$ will have a higher pressure than $\text{Ne}$ . However, the density of $\text{Ne}$ and $\text{He}$ are same.