What will become of the temperature of a sample of methane, if $1 \;L$ at $735\;torr$ and $25.0^oC$ is compressed to $1.8\; bar$ and a volume of $0.9\; L$ ?
$760\; torr$ $ =$ $ 1 \;atm$ $1\; atm$ $ =$ $ 1,013\; bar$
$P_1=\frac{735}{760}atm$ $P_2=\frac{1.8}{1.013}atm$ $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ $T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8\cdot 0.9\cdot 298\cdot 760}{1.013\cdot 735\cdot 1}=492.77K$
