Chapter 13, Problem 87GP

(a)

To determine

To Find: The density on a hot day.

Answer to Problem 87GP

The density at a temperature of ${38}^{\circ }C$ is $0.597\times {10}^3{\text{ kg/m}}^{\text{3}}$ .

Explanation of Solution

Given:

Initial temperature, $T_1=0°\text{C = 273 K}$ .

Final temperature, $T_2=38°\text{C = 311 K}$ .

Density at $0\text{°C, }{\rho }_1=0.68\times {10}^3{\text{ kg / m}}^{\text{3}}$ .

Formula used:

Ideal gas law formula is used to determine the density at increased temperature,

  $\begin{array}{l}P\times V=n\times R\times T\\ \text{n and R constant}\\ \frac{P\times V}{T}=\text{ constant}\end{array}$

Here, $P$ is the pressure,

  $V$ is the volume,

  $n$ is the number of molecules,

  $R$ is the universal gas constant,

  $T$ is the Temperature.

Write the expression for density.

  $\rho \text{ = }\frac{\text{Mass}}{\text{Volume}}$

  $\text{So, Volume }V\text{ = }\frac{\text{Mass}}{\text{Density}}$

Calculation:

Substitute the given values in the above equation,

  $\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}$

Pressure is same at both the temperature. Thus,

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

  $\frac{V_1}{V_2}=\frac{T_1}{T_2}$......... Equation (1)

Write the expression for the volume.

  $V\text{ = }\frac{\text{Mass}}{\text{Density}}$

Mass is same at both the temperatures. Thus,

  $V_1=\frac{1}{{\rho }_1}\text{ and }{V}_2=\frac{1}{{\rho }_2}$

  $\frac{V_1}{V_2}=\frac{{\rho }_2}{{\rho }_1}$......... Equation (2)

From equation (1) and equation (2)

  $\begin{array}{l}\frac{{\rho }_2}{{\rho }_1}=\frac{T_1}{T_2}\\ {\rho }_2=\frac{273}{311}\times 0.68\times {10}^3=0.597\times {10}^3{\text{ kg/m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the density at a temperature of $3{\text{8}}^{\text{o}}\text{C}$ is $0.597\times {10}^3{\text{ kg/m}}^{\text{3}}$ .

(b)

To determine

To Find: The percentage change in the density.

Answer to Problem 87GP

The percentage change in the density at both the temperature is $87.79\text{ \%}$ .

Explanation of Solution

Given:

Density at $0°\text{C}$ , ${\rho }_1=0.68\times {10}^3{\text{ kg / m}}^{\text{3}}$ .

Density at $38°\text{C}$ , ${\rho }_1=0.597\times {10}^3{\text{ kg / m}}^{\text{3}}$ .

Formula used:

The percentage change in the density at both the temperature is determined by the following temperature,

  $\text{\% change in density = }\frac{\text{Density at }38\text{°C}}{\text{Density at }0\text{°C}}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}\text{\% change in density }=\frac{0.597\times {10}^3}{0.68\times {10}^3}\times 100=0.8779\times 100\\ \text{\% change in density = }87.79\text{ \%}\end{array}$

Conclusion:

Thus,the percentage change in the density at both the temperature is $87.79\text{ \%}$ .