Answer
464 K
Work Step by Step
$n=13.7 \,g\,Cl_{2}\times\frac{1\,mol\,Cl_{2}}{70.906\,g\,Cl_{2}}=0.1932\,mol$
$R=0.08314\,bar\,L\,mol^{-1}K^{-1}$
$P=0.993\,bar$
$V=7.50\,L$
From the ideal gas law, we have
$ T=\frac{PV}{nR}=\frac{0.993\,bar\times7.50\,L}{0.1932\,mol\times0.08314\,bar\,L\,mol^{-1}K^{-1}}$
$=464\,K $