To find the best estimate for the volume \( V \) when the pressure \( P \) is \( 7.0 \times 10^3 \) pascals, we can use linear interpolation between the given data points.
The given data points are:
[tex]\[
\begin{array}{|c|c|}
\hline
P \, (\text{pascals}) & V \, (\text{liters}) \\
\hline
5.0 \times 10^3 & 6.0 \\
1.5 \times 10^4 & 3.0 \\
2.0 \times 10^4 & 2.0 \\
2.5 \times 10^4 & 1.5 \\
\hline
\end{array}
\][/tex]
We need to estimate the volume \( V \) when the pressure \( P \) is \( 7.0 \times 10^3 \) pascals.
1. Identify the interval in which \( 7.0 \times 10^3 \) pascals falls. That would be between \( 5.0 \times 10^3 \) and \( 1.5 \times 10^4 \) pascals.
2. Use the linear interpolation formula. For two points \((P_1, V_1)\) and \((P_2, V_2)\), the formula for interpolation is:
[tex]\[
V = V_1 + \frac{P - P_1}{P_2 - P_1} \times (V_2 - V_1)
\][/tex]
3. Substitute the known values into the formula:
[tex]\[
V = 6.0 + \frac{7.0 \times 10^3 - 5.0 \times 10^3}{1.5 \times 10^4 - 5.0 \times 10^3} \times (3.0 - 6.0)
\][/tex]
4. Simplify the values:
[tex]\[
V = 6.0 + \frac{2000}{10000} \times (-3.0)
\][/tex]
[tex]\[
V = 6.0 + 0.2 \times (-3.0)
\][/tex]
[tex]\[
V = 6.0 - 0.6
\][/tex]
[tex]\[
V = 5.4
\][/tex]
So, the best estimate for the value of \( V \) at \( P = 7.0 \times 10^3 \) pascals is \( 5.4 \) liters.
Therefore, the correct answer is [tex]\( \boxed{5.4 \text{ liters}} \)[/tex].
