Question

A given amount of oxygen gas has a volume of 25.0 L at a temperature of 37 °C and a pressure of 1.0 atm. At what temperature would this gas occupy a volume of 22.0 L at a pressure of 1.0 atm?

Answer 1

$V_1 = 25.0 \;L$

$T_1 = 37 \;°C + 273.15 = 310.15 \;K$

$p_1 = p_2 = 1.0 \;atm$

$V_2 = 22.0 \;L$

$T_2 = unknown$

$\frac{V_1 \times p_1}{T_1} = \frac{V_2 \times p_2}{T_2}$

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$T_2 = \frac{T_1 \times V_2}{V_1}$

$T_2 = \frac{310.15 \times 22.0}{25.0} = 272.93 \;K = -0.2 \;°C ≈ 0 \;°C$