At $25 \, ^\circ C$ temperature and $760 \, mm \, Hg$ pressure,the volume of a gas is $600 \, mL$. Calculate the pressure when the volume of this gas becomes $640 \, mL$ at $10 \, ^\circ C$ temperature. Use the ideal gas equation: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

(676.6 MM HG) Given:
$p_1 = 760 \, mm \, Hg$
$V_1 = 600 \, mL$
$T_1 = 25 + 273 = 298 \, K$
$V_2 = 640 \, mL$
$T_2 = 10 + 273 = 283 \, K$
Using the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$
$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}$
$p_2 = \frac{760 \times 600 \times 283}{298 \times 640}$
$p_2 = \frac{129048000}{190720} \approx 676.6 \, mm \, Hg$