Question

To what temperature must 30.0dm3 of helium at 25°C be cooled at 1 atm pressure for its volume to
be reduced to 1.0dm3 at same pressure?

Answer 1

Keeping the pressure constant,

$V\propto T$ or,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ And $V_1=30 \ dm^3,T_1=25\degree C=298 K$

$V_2=1 \ dm^3,T_2=?$

On substituting values,$\frac{30}{1}=\frac{298}{T_2}$ $\implies$ $T_2=9.93\ K=(9.93-273)\degree C=-263.07\degree C$