Question

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of gasoline occupies 0.003 80 m3. How many extra kilograms of gasoline would you receive if you bought 8.50 gal of gasoline at 0°C rather than at 21.7°C from a pump that is not temperature compensated?

Answer 1

To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we can use the formula for volume expansion and the density of gasoline.

Step-by-step explanation:

To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we need to calculate the difference in volume and then convert it to kilograms using the density of gasoline. The average coefficient of volume expansion is given as 9.60 x 10^-4 (°C)^-1. We can calculate the change in volume using the formula ΔV = V₀ * β * ΔT, where ΔV is the change in volume, V₀ is the initial volume, β is the coefficient of volume expansion, and ΔT is the temperature difference in Celsius. In this case, the initial volume is 0.00380 m³ and the temperature difference is 21.7 - 0 = 21.7 °C. Substituting the values, we get ΔV = 0.00380 * 9.60 x 10^-4 * 21.7. Now, to convert the change in volume to kilograms, we multiply it by the density of gasoline, which is 730 kg/m³. So, the extra kilograms of gasoline received is ΔV * 730.

Answer 2

Answer: 0.4911 kg

Step-by-step explanation:

We have the following data:

`\rho_(0\°C)= 730 kg/m^(3)` is the density of gasoline at
`0\°C`

`\beta=9.60(10)^(-4) \°C^(-1)` is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to
`m^(3)`:

`V=8.50 gal (0.00380 m^(3))/(1 gal)=0.0323 m^(3)` (1)

Knowing density is given by:
`\rho=(m)/(V)`, we can find the mass
`m_(1)` of 8.50 gallons:

`m_(1)=\rho_(0\°C)V`

`m_(1)=(730 kg/m^(3))(0.0323 m^(3))=23.579 kg` (2)

Now, we have to calculate the factor
`f` by which the volume of gasoline is increased with the temperature, which is given by:

`f=(1+\beta(T_(f)-T_(o)))` (3)

Where
`T_(o)=0\°C` is the initial temperature and
`T_(f)=21.7\°C` is the final temperature.

`f=(1+9.60(10)^(-4) \°C^(-1)(21.7\°C-0\°C))` (4)

`f=1.020832` (5)

With this, we can calculate the density of gasoline at
`21.7\°C`:

`\rho_(21.7\°C)=730 kg/m^(3) f=(730 kg/m^(3))(1.020832)`

`\rho_(21.7\°C)=745.207 kg/m^(3)` (6)

Now we can calculate the mass of gasoline at this temperature:

`m_(2)=\rho_(21.7\°C)V` (7)

`m_(2)=(745.207 kg/m^(3))(0.0323 m^(3))` (8)

`m_(2)=24.070 kg` (9)

And finally calculate the mass difference
`\Delta m`:

`\Delta m=m_(2)-m_(1)=24.070 kg-23.579 kg` (10)

`\Delta m=0.4911 kg` (11) This is the extra mass of gasoline