Chapter 9.5, Problem 44PPS

a.

To determine

To write an equation to represent the volume of a gas in terms of pressure and temperature.

Answer to Problem 44PPS

The required equation is $v=R\frac{t}{p},R\text{ is constant of variation}$

Explanation of Solution

Given information:

The given statement is that the volume of gas v varies inversely as the pressure p and directly as the temperature t .

The given relation is

  $v\propto \frac{1}{p},v\propto t$

The required equation is

  $v=R\frac{t}{p},R\text{ is constant of variation}$

b.

To determine

To write is the equation a direct, joint, inverse, or combined variation.

Answer to Problem 44PPS

The equation is combined variation.

Explanation of Solution

Given information:

The equation of variation is $v=R\frac{t}{p},R\text{ is constant of variation}$

The standard form of the equation

  $v=R\frac{t}{p},R\text{ is constant of variation}$

There are three variables in the equation v , p and t . the equation is of $y=k\frac{z}{x}$ type where $k=R$ , constant of the variation. Hence the equation is represents combined variation.

c.

To determine

To calculate the new pressure.

Answer to Problem 44PPS

The new pressure is 1.82 atmospheres

Explanation of Solution

Given information:

Initial volume is 8 liters, initial temperature is 275 kelvin and initial pressure is 1.25 atmospheres. New volume is 6 liters new temperature is 300 kelvin.

Calculation:

For the initial values

  $\begin{array}{l}{v}_1=R\frac{t_1}{p_1}\\ R=\frac{p_1{v}_1}{t_1}\left[\text{ simplify}\right]\\ \text{for new values}\\ v_2=R\frac{t_2}{p_2}\\ R=\frac{p_2{v}_2}{t_2}\left[\text{ simplify}\right]\\ \text{equate the value of }R\\ \frac{p_1{v}_2}{t_1}=\frac{p_2{v}_2}{t_2}\\ \frac{1.25\times 8}{275}=\frac{p_2\times 6}{300}\left[\text{substitute values of }{p}_1,v_1,t_1,t_2,v_2\right]\\ p_2=\frac{1.25\times 8\times 300}{275\times 6}\\ p_2\approx 1.82\end{array}$

Therefore,

The new pressure is 1.82 atmospheres