Chapter 10, Problem 26P

The density or gasoline is 7.30 × 10² kg/m³ at 0°C. Its average coefficient of volume expansion is 9.60 × 10⁻⁴(°C)⁻¹ and note that 1.00 gal = 0.003 80 m³. (a) Calculate the mass of 10.0 gal of gas at 0°C. (b) If 1.000 m³ of gasoline at 0°C is warmed by 20.0°C, calculate its new volume. (c) Using the answer to part (b), calculate the density of gasoline at 20.0°C. (d) Calculate the mass of 10.0 gal of gas at 20.0°C. (e) How many extra kilograms of gasoline would you get if you bought 10.0 gal of gasoline at 0°C rather than at 20.0°C from a pump that is not temperature compensated?

To determine

The mass of the gas.

Answer to Problem 26P

The mass of the gas is 27.7 kg.

Explanation of Solution

Given info: The density of gasoline ( ${\rho }_0$) is $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$. The coefficient of volume expansion ( $\beta$) is $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. Volume of the gas (V₀) is 10.0 gal. Temperature (T) is 0⁰C.

Formula to calculate the mass is,

$m_0={\rho }_0{V}_0$

Substitute $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$ for ${\rho }_0$ and 10.0 gal for V₀.

$\begin{array}{c}{m}_0=\left(7.30\times {10}^2{\text{kg/m}}^{\text{3}}\right)\left(\text{10}\text{.0}\text{gal}\right)\\ =\left(7.30\times {10}^2{\text{kg/m}}^{\text{3}}\right)\left(\text{10}\text{.0}\text{gal}\right)\left(\text{0}\text{.00380}{\text{m}}^{\text{3}}\text{/gal}\right)\\ =27.7\text{kg}\end{array}$

Conclusion:

The mass of the gas is 27.7 kg.

To determine

The new volume.

Answer to Problem 26P

The new volume is $1.02{\text{m}}^{\text{3}}$.

Explanation of Solution

Given info: The density of gasoline ( ${\rho }_0$) is $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$. The coefficient of volume expansion ( $\beta$) is $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. Volume of the gas (V₀) is 10.0 gal. Initial temperature (T₁) is 0⁰C. Final temperature (T₂) is 20⁰C.

Formula to calculate the volume is,

$V=V_0\left[1+\beta \left(T_2-T_1\right)\right]$

Substitute $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$ for $\beta$ and 10.0 gal for V₀.

$\begin{array}{c}V=\left(10.0\text{gal}\right)\left[1+\left(9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}\right)\left[\left({20}^{\text{ο}}\text{C}\right)-\left(0^{\text{ο}}\text{C}\right)\right]\right]\\ =\left(10.0\text{gal}\right)\left(\text{0}\text{.00380}{\text{m}}^{\text{3}}\text{/gal}\right)\left[1+\left(9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}\right)\left[\left({20}^{\text{ο}}\text{C}\right)-\left(0^{\text{ο}}\text{C}\right)\right]\right]\\ =1.02{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The new volume is $1.02{\text{m}}^{\text{3}}$.

To determine

The new density.

Answer to Problem 26P

The new density is $716{\text{kg/m}}^3$.

Explanation of Solution

Given info: The density of gasoline ( ${\rho }_0$) is $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$. The coefficient of volume expansion ( $\beta$) is $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. Volume of the gas (V₀) is 10.0 gal. Initial temperature (T₁) is 0⁰C. Final temperature (T₂) is 20⁰C.

Gasoline occupies $7.30\times {10}^2\text{kg}$ in a volume of $1{\text{m}}^{\text{3}}$. Therefore, the density is,

$\begin{array}{c}\rho =\frac{7.30\times {10}^2\text{kg}}{\text{1}\text{.02}{\text{m}}^{\text{3}}}\\ =716{\text{kg/m}}^3\end{array}$

Conclusion:

The new density is $716{\text{kg/m}}^3$.

To determine

The new mass.

Answer to Problem 26P

The new mass is 27.2 kg.

Explanation of Solution

Given info: The density of gasoline ( ${\rho }_0$) is $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$. The coefficient of volume expansion ( $\beta$) is $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. Volume of the gas (V₀) is 10.0 gal. Initial temperature (T₁) is 0⁰C. Final temperature (T₂) is 20⁰C.

Formula to calculate the new mass is,

$m=\rho V_0$

Substitute $716{\text{kg/m}}^{\text{3}}$ for ${\rho }_0$ and 10.0 gal for V₀.

$\begin{array}{c}m=\left(716{\text{kg/m}}^{\text{3}}\right)\left(\text{10}\text{.0}\text{gal}\right)\\ =\left(716{\text{kg/m}}^{\text{3}}\right)\left(\text{10}\text{.0}\text{gal}\right)\left(\text{0}\text{.00380}{\text{m}}^{\text{3}}\text{/gal}\right)\\ =27.2\text{kg}\end{array}$

Conclusion:

The new mass is 27.2 kg.

To determine

The difference in mass.

Answer to Problem 26P

The difference in mass is 0.5 kg.

Explanation of Solution

Given info: The density of gasoline ( ${\rho }_0$) is $7.30\times {10}^2{\text{kg/m}}^{\text{3}}$. The coefficient of volume expansion ( $\beta$) is $9.60\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. Volume of the gas (V₀) is 10.0 gal. Initial temperature (T₁) is 0⁰C. Final temperature (T₂) is 20⁰C.

Formula to calculate the difference in mass is,

$\Delta m=m_0-m$

Substitute 27.7 kg for $m_0$ and 27.2 kg for m.

$\begin{array}{c}\Delta m=\left(27.7\text{kg}\right)-\left(27.2\text{kg}\right)\\ =0.5\text{kg}\end{array}$

Conclusion:

The difference in mass is 0.5 kg.