Answer
6.73 mmHg
Work Step by Step
$P_{1}=1\,atm=760\, mmHg$ (standard pressure)
$V_{1}=25.0\,L$
$T_{1}=273\,K$ (standard temperature)
$V_{2}= 2460\,L$
$T_{2}=(-35+273)K=238\,K$
According to combined gas law,
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (n is constant)
$\implies P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}=\frac{760\, mmHg\times25.0\,L\times238\,K}{273\,K\times2460\,L}$
$=6.73\, mmHg$
