Chapter 11, Problem 11.79FU

Interpretation Introduction

a.

To determine: If the statement comparing the pressure of helium and neon is true or false.

Explanation of Solution

As per the ideal gas equation:

$\begin{array}{l}\text{PV = nRT}\text{-(1)}\\ \text{P = pressure ; V = volume }\\ \text{n = number of moles = }\frac{\text{m (mass)}}{\text{M (molar mass)}}\\ \text{R = Gas constant = 0}\text{.0821 Latm/mol-K}\\ \text{T = Temperature}\\ \text{Equation (1) can be written as:}\\ \text{PV = }\frac{\text{m}}{\text{M}}\text{RT}\text{-(2)}\end{array}$

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.

Interpretation Introduction

b.

To determine:

If the statement comparing the densities of helium and neon is true or false

Explanation of Solution

We know that, 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.

Conclusion

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.