Thermodynamic derivatives give $T=(\partial U/\partial S)_V=2\lambda V^{-2/3}S$ and $P=-(\partial U/\partial V)_S=(2/3)\lambda V^{-5/3}S^2$. Eliminating S using $S=TV^{2/3}/(2\lambda)$ gives $P=T^2V^{-1/3}/(6\lambda)$. Hence $PV^{1/3}/T^2=1/(6\lambda)$, A.
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