Chapter 8.2, Problem 10P

To determine

To calculate: Volume of helium at $\text{27}°\text{C}$ and $\text{121 kPa}$ .

Answer to Problem 10P

Volume of helium at $\text{27}°\text{C}$ and $\text{121 kPa}$ is $20.62\text{ L}$ .

Explanation of Solution

Given information:

The following variation:

The volume of a given mass of a gas varies directly as its absolute temperature and inversely as its pressure.

When temperature is $\text{2}°\text{C}$ and pressure is $\text{99 kPa}$ , $\text{4 g}$ of helium occupies $\text{21}\text{.3 L}$ .

Calculation:

Let V be volume, t be absolute temperature and P be pressure. Then, V varies directly as t and inversely as P ,

  $V=k\frac{t}{P}$

where k is a constant.

Substituting $V=21.3\text{ L}$ , $t=2°\text{C =}\left(2+273\right)\text{K = 275 K}$ and $P=99\text{ kPa}$ ,

  $\begin{array}{l}V=k\frac{t}{P}[\text{Equation]}\\ 99=\frac{k\left(275\right)}{23.1}[\text{Substitute]}\\ k=\frac{99\times 23.1}{275}[\text{Multiply both sides by 23}\text{.1/275]}\\ k=8.316[\text{Simplify]}\end{array}$

Therefore, an equation of the given variation is $V=8.316\frac{t}{P}$

To find S when $t=27°\text{C}=\left(27+273\right)\text{K}=300\text{ K}$ and $P=121\text{ kPa}$ , substitute in $V=8.316\frac{t}{P}$ .

  $\begin{array}{l}V=8.316\frac{t}{P}[\text{Equation]}\\ V=\frac{8.316\left(300\right)}{121}[\text{Substitute]}\\ V=20.62[\text{Simplify]}\end{array}$