A tanker ship is filled with `2.25 xx10^(5) m^3` of gasoline at a refinery in southern Texas when the temperature is `17.2 ^@C` .When the ship arrives in New York City, the temperature is `1.3^@C` . If the coefficient of volume expansion for gasoline is `9.50xx10^(-4)//""^@C` , how much has the volume of the gasoline decreased when it is unloaded in New York ?

Options

A. `1.50 xx10^(-2)m^3`
B. `66.2m^3`
C. `1290 m^3`
D. `3400 m^3`

Answer

To solve the problem of how much the volume of gasoline has decreased when unloaded in New York, we can follow these steps:

### Step 1: Identify the initial conditions
- Initial volume of gasoline, \( V_i = 2.25 \times 10^5 \, m^3 \)
- Initial temperature, \( T_i = 17.2 \, ^\circ C \)
- Final temperature, \( T_f = 1.3 \, ^\circ C \)
- Coefficient of volume expansion for gasoline, \( \alpha_v = 9.50 \times 10^{-4} \, ^\circ C^{-1} \)

### Step 2: Calculate the change in temperature
The change in temperature, \( \Delta T \), can be calculated as:
\[ \Delta T = T_f - T_i = 1.3 - 17.2 = -15.9 \, ^\circ C \]

### Step 3: Use the formula for volume change
The formula for the change in volume due to temperature change is given by:
\[ \Delta V = V_i \times \alpha_v \times \Delta T \]

### Step 4: Substitute the values into the formula
Now, substituting the values we have:
\[ \Delta V = (2.25 \times 10^5 \, m^3) \times (9.50 \times 10^{-4} \, ^\circ C^{-1}) \times (-15.9 \, ^\circ C) \]

### Step 5: Perform the calculations
Calculating the above expression:
1. Calculate \( (2.25 \times 10^5) \times (9.50 \times 10^{-4}) \):
\[ = 2.25 \times 9.50 \times 10^{5-4} = 21.375 \times 10^1 = 213.75 \, m^3 \]
2. Now multiply by \( -15.9 \):
\[ \Delta V = 213.75 \times (-15.9) \approx -3391.875 \, m^3 \]

### Step 6: Interpret the result
The negative sign indicates a decrease in volume. Thus, the volume of gasoline has decreased by approximately \( 3391.875 \, m^3 \).

### Final Answer
The volume of gasoline has decreased by approximately \( 3391.88 \, m^3 \) when unloaded in New York.

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