= Solution
Taking a <derivative> of the logarithmic form gives a direct way to use the plot's base-ten vertical axis:
$$
\boxed{L\simeq R\ln10\,T^2\frac{d\log_{10}(p/\mathrm{Pa})}{dT}.}
$$
Alternatively, use two nearby readings and eliminate the unknown $p_0$:
$$
L\simeq R\ln10\,
\frac{\log_{10}(p_b/\mathrm{Pa})-\log_{10}(p_a/\mathrm{Pa})}{T_a^{-1}-T_b^{-1}}.
$$
Reading the original PDF graph approximately gives $\log_{10}(p/\mathrm{Pa})\simeq-0.6$ at $300\,\mathrm K$ and $\simeq3.8$ at $500\,\mathrm K$. These bracket the requested <temperature> and give an average <latent heat> near $400\,\mathrm K$:
$$
L\simeq(8.31\,\mathrm{J\,K^{-1}\,mol^{-1}})(2.3026)
\frac{4.4}{1/300-1/500}\,\mathrm K
\simeq6.3\times10^4\,\mathrm{J\,mol^{-1}}.
$$
Thus a suitable graph-based estimate is
$$
\boxed{L(400\,\mathrm K)\approx6\times10^4\,\mathrm{J\,mol^{-1}}\quad\text{(roughly }60\text{--}65\,\mathrm{kJ\,mol^{-1}}\text{).}}
$$
The graph does not warrant several significant figures; a local tangent or different reasonable nearby readings give a comparable estimate. The factor $\ln10$ is essential when using a base-ten logarithmic axis.
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