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zhenek [66]
3 years ago
7

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?
Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer: 0.4911 kg

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 \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{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

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Answer:

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Explanation:

From the question,

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make a the subject of the  equation

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Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

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a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

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t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

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Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state
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Answer:

a) W =400 kJ

b) W = 0 kJ

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Explanation:

<u>Given  </u>

<u><em>Process 1 ---> 2 </em></u>

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

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The relation of the process V = constant  

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The relation of the process V = constant

V3 = V2

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Volume of point (3) V3 = 4 m^3

<u>Process 3 ---> 1 </u>

The relation of the process PV = constant  

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Sketch the processes on the PV coordinates

The work for each process in kJ  

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The work is defined by  

W=\int\limits^a_b {x} \, dx

<em>a=V2</em>

<em>b=V1</em>

<em>x=P</em>

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P3 = P4 = 5 bar  

W=\int\limits^a_b {x} \, dx

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putting the value of a, b, x, dx in above integral

W=400 kJ

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x=1V^-1

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putting the value of a, b, x, dx in above integral

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