To estimate the value of [tex]\( V \)[/tex] at [tex]\( P = 7.0 \times 10^3 \)[/tex] pascals, we can use linear interpolation between the two closest data points that enclose the target pressure. Here are the given data points:
1. [tex]\( P = 5.0 \times 10^3 \)[/tex] pascals, [tex]\( V = 6.0 \)[/tex] liters
2. [tex]\( P = 1.5 \times 10^4 \)[/tex] pascals, [tex]\( V = 3.0 \)[/tex] liters
3. [tex]\( P = 2.0 \times 10^4 \)[/tex] pascals, [tex]\( V = 2.0 \)[/tex] liters
4. [tex]\( P = 2.5 \times 10^4 \)[/tex] pascals, [tex]\( V = 1.5 \)[/tex] liters
Since [tex]\( 7.0 \times 10^3 \)[/tex] pascals is between [tex]\( 5.0 \times 10^3 \)[/tex] and [tex]\( 1.5 \times 10^4 \)[/tex] pascals, we will use the first two data points for linear interpolation:
- [tex]\( (P_1, V_1) = (5.0 \times 10^3, 6.0) \)[/tex]
- [tex]\( (P_2, V_2) = (1.5 \times 10^4, 3.0) \)[/tex]
The formula for linear interpolation is:
[tex]\[ V = V_1 + \frac{(V_2 - V_1)}{(P_2 - P_1)} \cdot (P - P_1) \][/tex]
Substituting the values:
[tex]\[ V = 6.0 + \frac{(3.0 - 6.0)}{(1.5 \times 10^4 - 5.0 \times 10^3)} \cdot (7.0 \times 10^3 - 5.0 \times 10^3) \][/tex]
Simplifying the above expression:
[tex]\[ V = 6.0 + \frac{-3.0}{1.0 \times 10^4} \cdot 2.0 \times 10^3 \][/tex]
[tex]\[ V = 6.0 + (-0.3) \cdot 2 \][/tex]
[tex]\[ V = 6.0 - 0.6 \][/tex]
[tex]\[ V = 5.4 \][/tex]
Therefore, the best estimate for the value of [tex]\( V \)[/tex] at [tex]\( P = 7.0 \times 10^3 \)[/tex] pascals is [tex]\( 5.4 \)[/tex] liters.
Thus, the correct answer is:
B. 5.4 liters
