Answer
\begin{aligned}
P^{\prime}(303)& \approx 0.00265 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(313)& \approx 0.004145 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(323)& \approx 0.006295 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(333)& \approx 0.00931 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(343)& \approx 0.013435 \mathrm{atm} / \mathrm{K}
\end{aligned}
Work Step by Step
We use the symmetric difference quotient
$$ f'(a) =\frac{f(a+h)-f(a-h)}{2h}$$
and from the given table, we have
\begin{aligned}
P^{\prime}(303)& \approx \frac{P(313)-P(293)}{20}=\frac{0.0808-0.0278}{20}=0.00265 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(313)& \approx \frac{P(323)-P(303)}{20}=\frac{0.1311-0.0482}{20}=0.004145 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(323)& \approx \frac{P(333)-P(313)}{20}=\frac{0.2067-0.0808}{20}=0.006295 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(333)& \approx \frac{P(343)-P(323)}{20}=\frac{0.3173-0.1311}{20}=0.00931 \mathrm{atm} / \mathrm{K}\\
P^{\prime}(343)& \approx \frac{P(353)-P(333)}{20}=\frac{0.4754-0.2067}{20}=0.013435 \mathrm{atm} / \mathrm{K}
\end{aligned}
