Question

Hydrogen initially occupies 194 liters. If it changes to a volume of 73 L at 0° C and 760 mm Hg. If the initial temperature of the hydrogen was 30° C, what was its initial pressure of the gas in mm Hg? Round to two decimal places.

Answer 1

Let's denote the parameters of the gas in the initial condition by index 1, and in the final condition by index 2:

$V_1 = 194~\text{L},~t_1 = 30~\degree\text{C},~P_1 - ?~; \\ V_2 = 73~\text{L},~t_2 = 0~\degree\text{C},~P_2 = 760~\text{mmHg}.$

We are going to make use Combined gas law for the case when comparing the same substance under two different sets of conditions:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}.$

For the formula to be correct, the Celsius temperatures should be converted to absolute temperatures (in Kelvin):

$T = (\frac{t}{\degree\text{C}} + 273.15)~\text{K}; \\ T_1 = (30 + 273.15)~\text{K} = 303.15~\text{K}; \\ T_2 = (0 + 273.15)~\text{K} = 273.15~\text{K}.$

Solving the Combined gas law for the unknown initial pressure, and entering the numerical values,

$P_1 = P_2\frac{V_2}{T_2}\frac{T_1}{V_1} = 760~\text{mmHg}~*~\frac{73~\cancel{\text{L}}}{273.15~\cancel{\text{K}}}~*~\frac{303.15~\cancel{\text{K}}}{194~\cancel{\text{L}}} \approx 317.39~\text{mmHg}.$