Question

Dinitrogen oxide (N₂O) gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet gas occupied 130 mL at 21◦C when the atmospheric pressure was 750 Torr. What volume would the same amount of dry dinitrogen oxide have occupied if collected at 750 Torr and 21 ◦C? The vapor pressure of water is 18.65 Torr at 21◦C. Answer in units of mL.

Answer 1

Answer:

126.73 mL

Explanation:

The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:

750 = 18.85 + p

p = 750 - 18.85

p = 731.15 torr

It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that

$\frac{p1V1}{T1} = \frac{p2V2}{T2}$

Where p is the pressure, V is the volume, T is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:

p1V1 = p2V2, so:

731.15x130 = 750xV2

750V2 = 95049.5

V2 = 126.73 mL