Chapter 19, Problem 62PQ

(a)

To determine

The density of gasoline on a very hot day ($T=95°\text{F}$).

Answer to Problem 62PQ

The density of gasoline on a very hot day ($T=95°\text{F}$) is $723.88{\text{kg/m}}^3$.

Explanation of Solution

Write the expression for the decreasing density of gasoline.

  ${\rho }_0=\frac{m}{V_0}$                                                                                                            (I)

Here, ${\rho }_0$ is the decreasing density of the gasoline, $m$ is the mass of the gasoline, and $V_0$ is the increasing volume of the gas.

Rearrange the equation (I) for $m$.

  $m={\rho }_0{V}_0$                                                                                                          (II)

Write the expression for the expansion of volume (Refer Equation 19.5).

  $\Delta V\approx \beta V_0\Delta T$                                                                                                   (III)

Here, $\Delta V$ is the change in volume, $V_0$ is the initial volume, $\beta$ is the coefficient of volume expansion, and $\Delta T$ is the change in temperature.

Write the expression for the original density of gasoline.

  $\rho =\frac{m}{V}$                                                                                                             (IV)

Here, $\rho$ is the density of the gasoline, $m$ is the mass of the gasoline, and $V$ is the change in volume of the gasoline container.

Conclusion:

Substitute equation (II) in the equation (IV) and replace $V_0+\Delta V$ for $V$.

  $\begin{array}{c}\rho =\frac{{\rho }_0{V}_0}{V}\\ =\frac{{\rho }_0{V}_0}{V_0+\Delta V}\\ =\frac{{\rho }_0}{1+\left(\frac{\Delta V}{V_0}\right)}\end{array}$

Substitute equation (III) in above equation.

  $\begin{array}{c}\rho =\frac{{\rho }_0}{1+\left(\frac{\beta V_0\Delta T}{V_0}\right)}\\ =\frac{{\rho }_0}{1+\beta \left(T_f-T_i\right)}\end{array}$                                                                                              (V)

Here, $T_i$ is the initial temperature and $T_f$ is the final temperature.

Convert initial temperature of the gas into degree Celsius.

  $\begin{array}{c}{T}_i=\left(60°\text{F}-32\right)\frac{5}{9}\\ =15.6°\text{C}\end{array}$

Convert final temperature of the gas into degree Celsius.

  $\begin{array}{c}{T}_f=\left(95°\text{F}-32\right)\frac{5}{9}\\ =35°\text{C}\end{array}$

Substitute $950\times {10}^{-6}°{\text{C}}^{-1}$ for $\beta$ (Refer Table 19.2), $737.22{\text{kg/m}}^3$ for ${\rho }_0$, $15.6°\text{C}$ for $T_i$, and $35°\text{C}$ for $T_f$ in equation (V) to find $\rho$.

  $\begin{array}{c}\rho =\frac{\left(737.22{\text{kg/m}}^3\right)}{\left[1+\left(950\times {10}^{-6}°{\text{C}}^{-1}\right)\left(35°\text{C}-15.6°\text{C}\right)\right]}\\ =723.88{\text{kg/m}}^3\end{array}$

Therefore, the density of gasoline on a very hot day ($T=95°\text{F}$) is $723.88{\text{kg/m}}^3$.

(b)

To determine

The mass of the gasoline purchased at a temperature of $60°\text{F}$.

Answer to Problem 62PQ

The mass of the gasoline purchased at a temperature of $60°\text{F}$ is $33.5\text{kg}$.

Explanation of Solution

Rearrange the equation (IV) from part (a) for $m$.

  $m=\rho V$                                                                                                      (VI)

Conclusion:

Substitute $737.22{\text{kg/m}}^3$ for $\rho$ (at $60°\text{F}$) and $12.0\text{gal}$ for $V$ in equation (VI) to find $m$.

      $\begin{array}{c}m=\left(737.22{\text{kg/m}}^3\right)\left(12.0\text{gal}\right)\left(\frac{0.003785{\text{m}}^3}{1\text{gal}}\right)\\ =33.5\text{kg}\end{array}$

Therefore, the mass of the gasoline purchased at a temperature of $60°\text{F}$ is $33.5\text{kg}$.

(c)

To determine

The mass of the gasoline purchased at a temperature of $95°\text{F}$.

Answer to Problem 62PQ

The mass of the gasoline purchased at a temperature of $95°\text{F}$ is $32.9\text{kg}$.

Explanation of Solution

Rearrange the equation (IV) from part (a) for $m$.

  $m=\rho V$

Conclusion:

Substitute $723.88{\text{kg/m}}^3$ for $\rho$ (at $95°\text{F}$) and $12.0\text{gal}$ for $V$ in above equation to find $m$.

      $\begin{array}{c}m=\left(723.88{\text{kg/m}}^3\right)\left(12.0\text{gal}\right)\left(\frac{0.003785{\text{m}}^3}{1\text{gal}}\right)\\ =32.9\text{kg}\end{array}$

Therefore, the mass of the gasoline purchased at a temperature of $95°\text{F}$ is $32.9\text{kg}$.

(d)

To determine

The amount of money did a consumer lose by buying gasoline on a very hot day.

Answer to Problem 62PQ

The amount of money did a consumer lose by buying gasoline on a very hot day is $54\text{cents}$.

Explanation of Solution

Since the consumer spent $30 for the 12 gallons of fuel, but on the hot day received 0.6 kg less compared to the 33.5 kg expected or 1.8% less.

Conclusion:

Hence, they lost about 1.8% of the $30 gas bill about $54\text{cents}$.

Therefore, the amount of money did a consumer lose by buying gasoline on a very hot day is $54\text{cents}$.