Seawater contains around 300 μM O2 when it is saturated. How much oxygen gas will be dissolved in 1 L of seawater assuming standard temperature and pressure? What if the temperature was raised 40 degrees celsius? Does temperature play a big role in the volume of gas?
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At standard temperature and pressure 1 mole of a gas occupies 22.4 lt.
Given concentration = 300 x 10^-6 M
and volume of seawater = 1 lt.
Thus, the amount of oxygen in 1 lt seawater = 300 x10^-6 moles/lt x 1 lt
= 300 x 1^-6 moles x 22/4 lt/mole = 6720 x10^-6 lt = 6.72 ml
When the temperature is changed to 40 degree celsius or 313 K, we can use Charles' law to determine the change in gas volume.
According to Charles' Law: V1/T1 = V2/T2 at constant pressure
where T1 is standard temperature (273 K), V1 is volume at standard conditions (6.72 ml, calculated earlier), T2 is new temperature (313 K) and V2 is the desired volume:
Thus, V2 = V1/T1 X T2 = 6.72/273 x 313 = 7.705 ml
Temperature plays a big role in volume of gas (or any matter), as temperature increases gas molecules attain higher energy and thus greater volume. In case of solids, a temperature increase results in expansion.
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